You and simplyunlucky are my favorite yugitubers, I love watching all your new and old videos and I really like how you don't split these huge pack openings into random parts. I am on my iPhone so it's sometimes annoying to search and find the next part lol but thanks to you I don't need to worry! (:
Congrats, Cyberknight! I was wondering what your 1000th video was going to be! I hope that you continue to do this well on TH-cam, and most of all to have fun.
Love the video and congrats with the 40,000 subs, and all the vids. For the contest idea could you do something like a box giveaway on a lotary. but still congrats
Cyberknight I dig your channel aside from the awesome pack openings you don't come off obnoxious like most of the YGO channels, you don't beat us down with your opinions or spend episode after episode complaining about decks. Keep up the good work.
big fan love these videos keep up the good work 2 things 1) wat is the big deal about slushy 2)if ur only after holos y not just buy them online and save ur money just always been curious about that thanx
Your probably my favorite person on TH-cam that does yugioh because of you I got back into yugioh and got a couple packs and a tin and now I play again. ps great opening
Hey CK, great vid!! Listen, you mention wanting to run Ninjas in this video; a friend of mine ran Inzektor/Ninja for awhile.....You've got a Zektkaliber and Exa-Stag just right here in this opening if you wanted to give it a shot.
Tachyon Galaxy box!! Cyberknight you're a beast! All time favorite yugioh commentator! Could you make a harpie deck when the support comes out because i have one made on dueling network but im not sure i run them to their full potential.
DIMENSIONAL ANALYSIS We have seen that many quantities are denoted by specifying both a number and a unit. For example, the distance to the nearest telephone may be 8 meters, or the speed of a car might be 25 meters/second. Each quantity, according to its physical nature, requires a certain type of unit. Distance must be measured in a length unit such as meters, feet, or miles, and a time unit will not do. Likewise, the speed of an object must be specified as a length unit divided by a time unit. In physics, the term dimension is used to refer to the physical nature of a quantity and the type of unit used to specify it. Distance has the di- mension of length, which is symbolized as [L], while speed has the dimensions of length [L] divided by time [T], or [L/T]. Many physical quantities can be expressed in terms of a combination of fundamental dimensions such as length [L], time [T], and mass [M]. Later on, we will encounter certain other quantities, such as temperature, which are also funda- mental. A fundamental quantity like temperature cannot be expressed as a combination of the dimensions of length, time, mass, or any other fundamental dimension. Dimensional analysis is used to check mathematical relations for the consistency of their dimensions. As an illustration, consider a car that starts from rest and accelerates to a speed v in a time t. Suppose we wish to calculate the distance x traveled by the car but are not sure whether the correct relation is or We can decide by checking the quantities on both sides of the equals sign to see whether they have the same x 1 2vt.x 1 2vt 2 *Weight and mass are different concepts, and the relationship between them will be discussed in Section 4.7. 2762T_ch01_001-027.qxd 4/14/08 7:14 PM Page 5 dimensions. If the dimensions are not the same, the relation is incorrect. For , we use the dimensions for distance [L], time [T], and speed [L/T] in the following way: Dimensions Dimensions cancel just like algebraic quantities, and pure numerical factors like have no dimensions, so they can be ignored. The dimension on the left of the equals sign does not match those on the right, so the relation cannot be correct. On the other hand, ap- plying dimensional analysis to we find that Dimensions The dimension on the left of the equals sign matches that on the right, so this relation is dimensionally correct. If we know that one of our two choices is the right one, then is it. In the absence of such knowledge, however, dimensional analysis cannot identify the correct relation. It can only identify which choices may be correct, since it does not account for numerical factors like or for the manner in which an equation was derived from physics principles. 1 2 x 1 2vt [L]L T [T] [L] x 1 2 vt x 1 2vt, x 1 2vt2 1 2 [L]L T [T]2 [L][T] x 1 2 vt2 x 1 2vt 2 6 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS ( The answers are given at the end of the book. ) 1. (a) Is it possible for two quantities to have the same dimensions but different units? (b) Is it possible for two quantities to have the same units but different dimensions? 2. You can always add two numbers that have the same units (such as 6 meters 3 meters). Can you always add two numbers that have the same dimensions, such as two numbers that have the dimensions of length [L]? 3. The following table lists four variables, along with their units: Variable Units x Meters (m) v Meters per second (m/s) t Seconds (s) a Meters per second squared (m/s2) These variables appear in the following equations, along with a few numbers that have no units. In which of the equations are the units on the left side of the equals sign consistent with the units on the right side? (a) (d) (b) (e) (c) (f) 4. In the equation y
cn at 2 you wish to determine the integer value (1, 2, etc.) of the ex- ponent n . The dimensions of y, a, and t are known. It is also known that c has no dimen- sions. Can dimensional analysis be used to determine n ? t B2x a v at v3 2ax2x vt 1 2at2 v at 1 2at3x vt CHECK YOUR UNDERSTANDING Scientists use mathematics to help them describe how the physical universe works, and trigonometry is an important branch of mathematics. Three trigonometric functions are utilized throughout this text. They are the sine, the cosine, and the tangent of the angle TRIGONOMETRY 1.4 Problem-solving insight You can check for errors that may have arisen during algebraic manipulations by performing a dimensional analysis on the final expression. 2762T_ch01_001-027.qxd 4/14/08 7:14 PM Page 6 1.4 TRIGONOMETRY 7 (Greek theta), abbreviated as sin , cos , and tan , respectively. These functions are de- fined below in terms of the symbols given along with the right triangle in Figure 1.5. DEFINITION OF SIN , COS , AND TAN (1.1) (1.2) (1.3) h length of the hypotenuse of a right triangle ho length of the side opposite the angle ha length of the side adjacent to the angle tan
ho ha cos
ha h sin
ho h The sine, cosine, or tangent may be used in calculations such as that in Example 4, de- pending on which side of the triangle has a known value and which side is asked for. However, the choice of which side of the triangle to label ho (opposite) and which to label ha (adjacent) can be made only after the angle is identified. Often the values for two sides of the right triangle in Figure 1.5 are available, and the value of the angle is unknown. The concept of inverse trigonometric functions plays an important role in such situations. Equations 1.4-1.6 give the inverse sine, inverse cosine, and inverse tangent in terms of the symbols used in the drawing. For instance, Equation 1.4 is read as “ equals the angle whose sine is ho/h.” (1.4) (1.5) (1.6)
tan1 ho ha
cos1 ha h
sin1 ho h The sine, cosine, and tangent of an angle are numbers without units, because each is the ratio of the lengths of two sides of a right triangle. Example 4 illustrates a typical applica- tion of Equation 1.3. On a sunny day, a tall building casts a shadow that is 67.2 m long. The angle between the sun’s rays and the ground is 50.0°, as Figure 1.6 shows. Determine the height of the building. Reasoning We want to find the height of the building. Therefore, we begin with the colored right triangle in Figure 1.6 and identify the height as the length ho of the side opposite the an- gle . The length of the shadow is the length ha of the side that is adjacent to the angle . The ratio of the length of the opposite side to the length of the adjacent side is the tangent of the an- gle , which can be used to find the height of the building. Solution We use the tangent function in the following way, with 50.0 and ha 67.2 m: (1.3) The value of tan 50.0 is found by using a calculator. 80.0 m ho ha tan (67.2 m)(tan 50.0) (67.2 m)(1.19) tan
ho ha Example 4 Using Trigonometric Functions
90° h = hypotenuse θ ho = length of side opposite the angle ha = length of side adjacent to the angle θ θ = 50.0° θ = 67.2 mh a ho Figure 1.5 A right triangle. Figure 1.6 From a value for the angle and the length ha of the shadow, the height ho of the building can be found using trigonometry. Problem-solving insight 2762T_ch01_001-027.qxd 4/14/08 7:14 PM Page 7 The use of 1 as an exponent in Equations 1.4-1.6 does not mean “take the reciprocal.” For instance, (ho/ha) does not equal 1/tan (ho/ha). Another way to express the in- verse trigonometric functions is to use arc sin, arc cos, and arc tan instead of and tan Example 5 illustrates the use of an inverse trigonometric function. 1 cos. 1, sin1, tan1 8 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS A lakefront drops off gradually at an angle , as Figure 1.7 indicates. For safety reasons, it is necessary to know how deep the lake is at various distances from the shore. To provide some information about the depth, a lifeguard rows straight out from the shore a distance of 14.0 m and drops a weighted fishing line. By measuring the length of the line, the lifeguard determines the depth to be 2.25 m. (a) What is the value of ? (b) What would be the depth d of the lake at a distance of 22.0 m from the shore? Reasoning Near the shore, the lengths of the opposite and adjacent sides of the right triangle in Figure 1.7 are ho 2.25 m and ha 14.0 m, relative to the angle . Having made this iden- tification, we can use the inverse tangent to find the angle in part (a). For part (b) the opposite and adjacent sides farther from the shore become ho d and ha 22.0 m. With the value for obtained in part (a), the tangent function can be used to find the unknown depth. Considering the way in which the lake bottom drops off in Figure 1.7, we expect the unknown depth to be greater than 2.25 m. Solution (a) Using the inverse tangent given in Equation 1.6, we find that (b) With 9.13°, the tangent function given in Equation 1.3 can be used to find the unknown depth farther from the shore, where ho d and ha 22.0 m. Since tan ho/ha, it follows that which is greater than 2.25 m, as expected. 3.54 m d (22.0 m)(tan 9.13) ho ha tan 9.13
tan1 ho ha tan1 2.25 m 14.0 m Example 5 Using Inverse Trigonometric Functions
14.0 m 2.25 m d 22.0 m θ Figure 1.7 If the distance from the shore and the depth of the water at any one point are known, the angle can be found with the aid of trigonometry. Knowing the value of is useful, be- cause then the depth d at another point can be determined. The right triangle in Figure 1.5 provides the basis for defining the various trigonometric functions according to Equations 1.1-1.3. These functions always involve an angle and two sides of the triangle. There is also a relationship among the lengths of the three sides of a right triangle. This relationship is known as the Pythagorean theorem and is used often in this text. PYTHAGOREAN THEOREM The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides: (1.7)h 2 ho 2 ha 2 2762T_ch01_001-027.qxd 4/14/08 7:14 PM Page 8 1.5 SCALARS AND VECTORS 9 The volume of water in a swimming pool might be 50 cubic meters, or the winning time of a race could be 11.3 seconds. In cases like these, only the size of the numbers mat- ters. In other words, how much volume or time is there? The 50 specifies the amount of water in units of cubic meters, while the 11.3 specifies the amount of time in seconds. Volume and time are examples of scalar quantities. A scalar quantity is one that can be described with a single number (including any units) giving its size or magnitude. Some other common scalars are temperature (e.g., 20 C) and mass (e.g., 85 kg). While many quantities in physics are scalars, there are also many that are not, and for these quantities the magnitude tells only part of the story. Consider Figure 1.8, which de- picts a car that has moved 2 km along a straight line from start to finish. When describing the motion, it is incomplete to say that “the car moved a distance of 2 km.” This statement would indicate only that the car ends up somewhere on a circle whose center is at the start- ing point and whose radius is 2 km. A complete description must include the direction along with the distance, as in the statement “the car moved a distance of 2 km in a direc- tion 30 north of east.” A quantity that deals inherently with both magnitude and direction is called a vector quantity. Because direction is an important characteristic of vectors, ar- rows are used to represent them; the direction of the arrow gives the direction of the vec- tor. The colored arrow in Figure 1.8, for example, is called the displacement vector, because it shows how the car is displaced from its starting point. Chapter 2 discusses this particular vector. The length of the arrow in Figure 1.8 represents the magnitude of the displacement vector. If the car had moved 4 km instead of 2 km from the starting point, the arrow would have been drawn twice as long. By convention, the length of a vector arrow is propor- tional to the magnitude of the vector. In physics there are many important kinds of vectors, and the practice of using the length of an arrow to represent the magnitude of a vector applies to each of them. All forces, for instance, are vectors. In common usage a force is a push or a pull, and the di- rection in which a force acts is just as important as the strength or magnitude of the force. The magnitude of a force is measured in SI units called newtons (N). An arrow representing a force of 20 newtons is drawn twice as long as one representing a force of 10 newtons. The fundamental distinction between scalars and vectors is the characteristic of direc- tion. Vectors have it, and scalars do not. Conceptual Example 6 helps to clarify this dis- tinction and explains what is meant by the “direction” of a vector. SCALARS AND VECTORS 1.5 N S WE 30.0° Start Finish 2 km Figure 1.8 A vector quantity has a magnitude and a direction. The colored arrow in this drawing represents a dis- placement vector. There are places where the temperature is 20 C at one time of the year and 20 C at an- other time. Do the plus and minus signs that signify positive and negative temperatures imply that temperature is a vector quantity? (a) Yes (b) No Reasoning A hallmark of a vector is that there is both a magnitude and a physical direction associated with it, such as 20 meters due east or 20 meters due west. Answer (a) is incorrect. The plus and minus signs associated with 20 C and 20C do not convey a physical direction, such as due east or due west. Therefore, temperature cannot be a vector quantity. Answer (b) is correct. On a thermometer, the algebraic signs simply mean that the tempera- ture is a number less than or greater than zero on the temperature scale being used and have nothing to do with east, west, or any other physical direction. Temperature, then, is not a vec- tor. It is a scalar, and scalars can sometimes be negative. Conceptual Example 6 Vectors, Scalars, and the Role of Plus and Minus Signs
Often, for the sake of convenience, quantities such as volume, time, displacement, velocity, and force are represented in physics by symbols. In this text, we write vectors in Each of these racers in the Tour de France has a velocity as they approach the finish line on the Champs Élysées in Paris. Velocity is an example of a vector quantity, because it has a magnitude (the speed of the racer) and a direction (toward the finish line). (Owen Franken/Corbis Images) 2762T_ch01_001-027.qxd 4/14/08 7:14 PM Page 9 boldface symbols (this is boldface) with arrows above them* and write scalars in italic symbols (this is italic). Thus, a displacement vector is written as “ , due east,” where the is a boldface symbol. By itself, however, separated from the direction, the mag- nitude of this vector is a scalar quantity. Therefore, the magnitude is written as “A750 m,” where the A is an italic symbol without an arrow. A B A B 750 m 10 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS *Vectors are also sometimes written in other texts as boldface symbols without arrows above them. ADDITION Often it is necessary to add one vector to another, and the process of addition must take into account both the magnitude and the direction of the vectors. The simplest situa- tion occurs when the vectors point along the same direction-that is, when they are colin- ear, as in Figure 1.9. Here, a car first moves along a straight line, with a displacement vector of 275 m, due east. Then, the car moves again in the same direction, with a displacement vector of 125 m, due east. These two vectors add to give the total dis- placement vector , which would apply if the car had moved from start to finish in one step. The symbol is used because the total vector is often called the resultant vector. With the tail of the second arrow located at the head of the first arrow, the two lengths sim- ply add to give the length of the total displacement. This kind of vector addition is identi- cal to the familiar addition of two scalar numbers (2 3 5) and can be carried out here only because the vectors point along the same direction. In such cases we add the individ- ual magnitudes to get the magnitude of the total, knowing in advance what the direction must be. Formally, the addition is written as follows: Perpendicular vectors are frequently encountered, and Figure 1.10 indicates how they can be added. This figure applies to a car that first travels with a displacement vec- tor of 275 m, due east, and then with a displacement vector of 125 m, due north. The two vectors add to give a resultant displacement vector . Once again, the vectors to be added are arranged in a tail-to-head fashion, and the resultant vector points from the tail of the first to the head of the last vector added. The resultant displacement is given by the vector equation The addition in this equation cannot be carried out by writing R 275 m 125 m, be- cause the vectors have different directions. Instead, we take advantage of the fact that the triangle in Figure 1.10 is a right triangle and use the Pythagorean theorem (Equation 1.7). According to this theorem, the magnitude of is The angle in Figure 1.10 gives the direction of the resultant vector. Since the lengths of all three sides of the right triangle are now known, either sin , cos , or tan can be used R v(275 m)2 (125 m)2 302 m R B R B A B B B R B B B A B R B 275 m, due east 125 m, due east 400 m, due east R B A B B B R B R B B B A B VECTOR ADDITION AND SUBTRACTION 1.6 ( The answer is given at the end of the book. ) 5. Which of the following statements, if any, involves a vector? (a) I walked 2 miles along the beach. (b) I walked 2 miles due north along the beach. (c) I jumped off a cliff and hit the water traveling at 17 miles per hour. (d) I jumped off a cliff and hit the water traveling straight down at a speed of 17 miles per hour. (e) My bank account shows a negative balance of 25 dollars. CHECK YOUR UNDERSTANDING N S WE Tail-to-head Start Finish A R B N S WE Tail-to-head Start R B Finish 90° θ A Figure 1.9 Two colinear displacement vectors and add to give the resultant displacement vector . R B B B A B Figure 1.10 The addition of two per- pendicular displacement vectors and gives the resultant vector . R B B B A B 2762T_ch01_001-027.qxd 4/14/08 7:15 PM Page 10 1.6 VECTOR ADDITION AND SUBTRACTION 11 to determine . Noting that tan B/A and using the inverse trigonometric function, we find that: Thus, the resultant displacement of the car has a magnitude of 302 m and points north of east at an angle of 24.4. This displacement would bring the car from the start to the finish in Figure 1.10 in a single straight-line step. When two vectors to be added are not perpendicular, the tail-to-head arrangement does not lead to a right triangle, and the Pythagorean theorem cannot be used. Figure 1.11a illustrates such a case for a car that moves with a displacement of 275 m, due east, and then with a displacement of 125 m, in a direction 55.0° north of west. As usual, the re- sultant displacement vector is directed from the tail of the first to the head of the last vector added. The vector addition is still given according to However, the magnitude of is not RA B, because the vectors and do not have the same direction, and neither is it , because the vectors are not perpendicular, so the Pythagorean theorem does not apply. Some other means must be used to find the magnitude and direction of the resultant vector. One approach uses a graphical technique. In this method, a diagram is constructed in which the arrows are drawn tail to head. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor, perhaps). Then, the length of the arrow representing the resultant vector is measured with a ruler. This length is con- verted to the magnitude of the resultant vector by using the scale factor with which the drawing is constructed. In Figure 1.11b, for example, a scale of one centimeter of arrow length for each 10.0 m of displacement is used, and it can be seen that the length of the ar- row representing is 22.8 cm. Since each centimeter corresponds to 10.0 m of displace- ment, the magnitude of is 228 m. The angle , which gives the direction of , can be measured with a protractor to be 26.7° north of east. R B R B R B R 2A2 B2 B B A B R B R B A B B B R B B B A B
tan1 B A tan1 125 m 275 m 24.4 SUBTRACTION The subtraction of one vector from another is carried out in a way that depends on the following fact. When a vector is multiplied by 1, the magnitude of the vector remains the same, but the direction of the vector is reversed. Conceptual Example 7 illustrates the meaning of this statement. Finish Start Tail-to-head 55.0° 55.0° ( ) a ( ) b 22.8 cm 4 8 12 200 24 cm 16 28 N S WE θ θ A R B A R B Figure 1.11 (a) The two displacement vectors and are neither colinear nor perpendicular but even so they add to give the resultant vector . (b) In one method for adding them together, a graphical technique is used. R B B B A B Consider two vectors described as follows: 1. A woman climbs 1.2 m up a ladder, so that her displacement vector is 1.2 m, upward along the ladder, as in Figure 1.12a. 2. A man is pushing with 450 N of force on his stalled car, trying to move it eastward. The force vector that he applies to the car is 450 N, due east, as in Figure 1.13a. What are the physical meanings of the vectors and ? (a) points upward along the ladder and has a magnitude of 1.2 m; points due east and has a magnitude of 450 N. (b) points downward along the ladder and has a magnitude of 1.2 m; points due west and has a magnitude of 450 N. (c) points downward along the ladder and has a magnitude of 1.2 m; points due west and has a magnitude of 450 N. Reasoning A displacement vector of is (1) . The presence of the (1) factor reverses the direction of the vector, but does not change its magnitude. Similarly, a force vector of has the same magnitude as the vector but has the opposite direction. Answer (a) and (b) are incorrect. While scalars can sometimes be negative, magnitudes of vectors are never negative. F B F B D B D B F B D B F B D B F B D B F B D B F B D B Conceptual Example 7 Multiplying a Vector by 1 (a) D (b) -D Figure 1.12 (a) The displacement vec- tor for a woman climbing 1.2 m up a ladder is . (b) The displacement vec- tor for a woman climbing 1.2 m down a ladder is . D B D B 2762T_ch01_001-027.qxd 4/14/08 10:12 PM Page 11 Answer (c) is correct. The vectors and have the same magnitudes as and , but point in the opposite directions, as indicated in Figures 1.12b and 1.13b. Related Homework: Problem 33 F B D B F B D B
In practice, vector subtraction is carried out exactly like vector addition, except that one of the vectors added is multiplied by a scalar factor of 1. To see why, look at the two vectors and in Figure 1.14a. These vectors add together to give a third vector , according to . Therefore, we can calculate vector as , which is an example of vector subtraction. However, we can also write this result as ( ) and treat it as vector addition. Figure 1.14b shows how to calculate vec- tor by adding the vectors and . Notice that vectors and are arranged tail to head and that any suitable method of vector addition can be employed to determine . A B B B C B B B C B A B B B C B A B B B C B A B A B B B A B C B C B B B A B ( The answers are given at the end of the book .) 6. Two vectors and are added together to give a resultant vector : . The magnitudes of and are 3 m and 8 m, respectively, but the vectors can have any orientation. What is (a) the maximum possible value and (b) the minimum possible value for the magnitude of ? 7. Can two nonzero perpendicular vectors be added together so their sum is zero? 8. Can three or more vectors with unequal magnitudes be added together so their sum is zero? 9. In preparation for this question, review Conceptual Example 7. Vectors and satisfy the vector equation 0. (a) How does the magnitude of compare with the magnitude of ? (b) How does the direction of compare with the direction of ? 10. Vectors , , and satisfy the vector equation , and their magnitudes are related by the scalar equation A 2 B 2 C 2. How is vector oriented with respect to vector ? 11. Vectors , , and satisfy the vector equation , and their magnitudes are related by the scalar equation A B C . How is vector oriented with respect to vector ? B B A B C B B B A B C B B B A B B B A B C B B B A B C B B B A B A B B B A B B B B B A B B B A B R B B B A B B B A B R B R B B B A B CHECK YOUR UNDERSTANDING A B C = A + B ( ) a B- C Tail-to-head ( ) b A = C - B Figure 1.14 (a) Vector addition according to . (b) Vector subtraction according to ( ). B B C B B B C B A B B B A B C B VECTOR COMPONENTS Suppose a car moves along a straight line from start to finish, as in Figure 1.15, the corresponding displacement vector being . The magnitude and direction of the vector give the distance and direction traveled along the straight line. However, the car could also arrive at the finish point by first moving due east, turning through 90°, and then moving due north. This alternative path is shown in the drawing and is associated with the two dis- placement vectors and . The vectors and are called the x vector component and the y vector component of . Vector components are very important in physics and have two basic features that are apparent in Figure 1.15. One is that the components add together to equal the original vector:
The components and , when added vectorially, convey exactly the same meaning as does the original vector : they indicate how the finish point is displaced relative to the starting point. The other feature of vector components that is apparent in Figure 1.15 is that and are not just any two vectors that add together to give the original vector : they are perpen- dicular vectors. This perpendicularity is a valuable characteristic, as we will soon see. Any type of vector may be expressed in terms of its components, in a way similar to that illustrated for the displacement vector in Figure 1.15. Figure 1.16 shows an arbitrary r B y Bx B r B y Bx B y Bx Br B r B y Bx By Bx B r Br B THE COMPONENTS OF A VECTOR 1.7 Figure 1.13 (a) The force vector for a man pushing on a car with 450 N of force in a direction due east is . (b) The force vector for a man pushing on a car with 450 N of force in a direc- tion due west is . F B F B (a) (b) F -F B B Start Finish 90° N S WE x yr Figure 1.15 The displacement vector and its vector components and . y Bx B r B 12 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS 2762T_ch01_001-027.qxd 4/14/08 7:15 PM Page 12 1.7 THE COMPONENTS OF A VECTOR 13 DEFINITION OF VECTOR COMPONENTS In two dimensions, the vector components of a vector are two perpendicular vectors x and y that are parallel to the x and y axes, respectively, and add together vectori- ally so that xy .A B A B A B A B A B A B In general, the components of any vector can be used in place of the vector itself in any calculation where it is convenient to do so (Problem-solving insight). The values calcu- lated for vector components depend on the orientation of the vector relative to the axes used as a reference. Figure 1.18 illustrates this fact for a vector by showing two sets of axes, one set being rotated clockwise relative to the other. With respect to the black axes, vector has perpendicular vector components x and y; with respect to the colored ro- A B A B A B A B tated axes, vector has different vector components x and y . The choice of which set of components to use is purely a matter of convenience. A B A B A B Ay +y +x Ax A θ +y +x +y´ +x´ Ax A´y A´x A Ay SCALAR COMPONENTS It is often easier to work with the scalar components, Ax and Ay (note the italic symbols), rather than the vector components x and y. Scalar components are positive or negative numbers (with units) that are defined as follows: The scalar component Ax has a magnitude equal to that of x and is given a positive sign if x points along the axis and a negative sign if x points along the axis. The scalar component Ay is defined in a similar manner. The following table shows an example of vector and scalar components: Vector Components Scalar Components Unit Vectors x 8 meters, directed along the x axis meters x (8 meters) y 10 meters, directed along the y axis meters y (10 meters) In this text, when we use the term “component,” we will be referring to a scalar compo- nent, unless otherwise indicated. Another method of expressing vector components is to use unit vectors. A unit vector is a vector that has a magnitude of 1, but no dimensions. We will use a caret (^) to distin- guish it from other vectors. Thus, is a dimensionless unit vector of length l that points in the positive x direction, and is a dimensionless unit vector of length l that points in the positive y direction. These unit vectors are illustrated in Figure 1.19. With the aid of unit vectors, the vector components of an arbitrary vector can be written as Ax and Ay , where Ax and Ay are its scalar components (see the drawing and the third column of the table above). The vector is then written as Ax Ay .
can you PLLLLEEEEASEEEE do these requests? 1: After tachyon galaxy comes out, update galaxy deck? 2: Update the heraldic beasts deck 3: make a heroic challenger deck or until the number hunters pack comes out with the new heroic cards 4: and in time, when the other packs come out can you make the barian emperors' deck profiles love your vids cyber, keep up the great work =D
It's funny that you mentioned the meta and X-Sabers; I'm actually working on making X-Sabers meta again. I think they're probably around tier 1.5 with the build I'm currently using.
i barely like video but this and everyone of your videos are alsome XD, i watch videos when i am doing homework so i like around 15-50 min videos i agree with cyber knight
Wow.... I just noticed, this video was uploaded on my birthday!!! This would have been an awesome present. I love watching unboxings and want to start one myself!!! I need a tripod though :/
Cyberknight i have a question concerning what sites to order my english booster boxes, specifically the new Judgment of the light thats comming out soon. can you please pm me or something to tell me what sites you order your booster boxes from?
You and simplyunlucky are my favorite yugitubers, I love watching all your new and old videos and I really like how you don't split these huge pack openings into random parts. I am on my iPhone so it's sometimes annoying to search and find the next part lol but thanks to you I don't need to worry! (:
Congrats, Cyberknight! I was wondering what your 1000th video was going to be! I hope that you continue to do this well on TH-cam, and most of all to have fun.
i like the old yugioh
Love the video and congrats with the 40,000 subs, and all the vids. For the contest idea could you do something like a box giveaway on a lotary. but still congrats
just shared, love these, keep them up! :D
"let's knock off one of the packs off the table" Lol, made my day.
Congrats For 1000 vids keep up the awesome work
Epic Pulls
Good job on your videos and good job on 1000th videos!
It's been a pleasure bro. Nice job. Bringing the game to the next level
Congrats cyber I've subscribed a week ago and I love your vids, keep up the good work
Nice pulls Cyberknight! You even got a Slushy!
Hopefully you get 200,000 subs by the end of the year!
Cyberknight 1000th video! Congratulations!!! :D
YEEESSSS! haha awesome awesome awesome! haha love these 100 pack openings! keep up the good work cyber :)
Cyberknight I dig your channel aside from the awesome pack openings you don't come off obnoxious like most of the YGO channels, you don't beat us down with your opinions or spend episode after episode complaining about decks. Keep up the good work.
Congrats to your 1000 vid milestone!
Amazing :) congrats again! Some great pulls too.
What makes this channel unique are Cyberknight's bright attitude, the Japanese openings, and dem swagbucks lol.
big fan love these videos keep up the good work 2 things 1) wat is the big deal about slushy 2)if ur only after holos y not just buy them online and save ur money just always been curious about that thanx
Your probably my favorite person on TH-cam that does yugioh because of you I got back into yugioh and got a couple packs and a tin and now I play again. ps great opening
Hey CK, great vid!! Listen, you mention wanting to run Ninjas in this video; a friend of mine ran Inzektor/Ninja for awhile.....You've got a Zektkaliber and Exa-Stag just right here in this opening if you wanted to give it a shot.
Cyberknight thx for the awesome content almonst every day, I don't play yu-gi-oh anymore but I still really enjoy watching your video :D
Congrats on the 1000th vid CK!
happy 1000 video i love this channel so much keep it up
Great pack openings. Also I remember Dragon Queen of Tragic Endings was used by Dark Signer Misty in Yu-Gi-Oh! 5D's.
CONGRATS ON 1,000 SUBSCRIBERS!!!!
I love when you do the 100 random pack opening video!!!!! and you got some nice cards!
watch it with 0,5 Speed =D sounds so drunk.
+YGC Kha Zeel - DAMN I LOVE YOU
Holy shit, he really does.
"Come on! Let's see some Ghost rares, Secret rares, CRAZINESS!" Hahaha!
Hilarious!
Cyperknight you Legend!!!! Happy 1000th videos :D
Congrats on the 1000th video!!!
OMG this is great. I am happy to see you doing this again
Tachyon Galaxy box!! Cyberknight you're a beast! All time favorite yugioh commentator! Could you make a harpie deck when the support comes out because i have one made on dueling network but im not sure i run them to their full potential.
thanks for doin this for us, i appriciate it. u have the best yugioh channel there is
Congrats on the 1000th Video!!!!!
Love the vid CyberKnight. Keep up the great work :)
Sorry cyber, I liked before even looking at the video. Congrats on the 1000 video! :D
Congrats on making 1000 videos!
Congrats on your 1000 vid
Ninja deck profile yeeeee!! :D wow the whole 43:27 watched :P keep it up knight :)
congratz on 1000 vids.
Congrats on 1000 cyber
the way youre opening those packs, i am not surprised you didnt get more holos from this opening haha
what was it you were saying about the madolche cards in the video?
You are a great you-tuber and im glad you started youtube keep it up
DIMENSIONAL ANALYSIS
We have seen that many quantities are denoted by specifying both a number and a unit. For example, the distance to the nearest telephone may be 8 meters, or the speed of a car might be 25 meters/second. Each quantity, according to its physical nature, requires a certain type of unit. Distance must be measured in a length unit such as meters, feet, or miles, and a time unit will not do. Likewise, the speed of an object must be specified as a length unit divided by a time unit. In physics, the term dimension is used to refer to the physical nature of a quantity and the type of unit used to specify it. Distance has the di- mension of length, which is symbolized as [L], while speed has the dimensions of length [L] divided by time [T], or [L/T]. Many physical quantities can be expressed in terms of a combination of fundamental dimensions such as length [L], time [T], and mass [M]. Later on, we will encounter certain other quantities, such as temperature, which are also funda- mental. A fundamental quantity like temperature cannot be expressed as a combination of the dimensions of length, time, mass, or any other fundamental dimension. Dimensional analysis is used to check mathematical relations for the consistency of their dimensions. As an illustration, consider a car that starts from rest and accelerates to a speed v in a time t. Suppose we wish to calculate the distance x traveled by the car but are not sure whether the correct relation is or We can decide by checking the quantities on both sides of the equals sign to see whether they have the same x 1 2vt.x 1 2vt 2
*Weight and mass are different concepts, and the relationship between them will be discussed in Section 4.7.
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dimensions. If the dimensions are not the same, the relation is incorrect. For , we use the dimensions for distance [L], time [T], and speed [L/T] in the following way:
Dimensions
Dimensions cancel just like algebraic quantities, and pure numerical factors like have no dimensions, so they can be ignored. The dimension on the left of the equals sign does not match those on the right, so the relation cannot be correct. On the other hand, ap- plying dimensional analysis to we find that
Dimensions
The dimension on the left of the equals sign matches that on the right, so this relation is dimensionally correct. If we know that one of our two choices is the right one, then is it. In the absence of such knowledge, however, dimensional analysis cannot identify the correct relation. It can only identify which choices may be correct, since it does not account for numerical factors like or for the manner in which an equation was derived from physics principles. 1 2 x 1 2vt [L]L T [T] [L] x 1 2 vt
x 1 2vt, x 1 2vt2
1 2
[L]L T [T]2 [L][T] x 1 2 vt2
x 1 2vt 2
6 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS
( The answers are given at the end of the book. )
1. (a) Is it possible for two quantities to have the same dimensions but different units? (b) Is it possible for two quantities to have the same units but different dimensions? 2. You can always add two numbers that have the same units (such as 6 meters 3 meters). Can you always add two numbers that have the same dimensions, such as two numbers that have the dimensions of length [L]?
3. The following table lists four variables, along with their units:
Variable Units
x
Meters (m) v Meters per second (m/s) t Seconds (s) a Meters per second squared (m/s2)
These variables appear in the following equations, along with a few numbers that have no units. In which of the equations are the units on the left side of the equals sign consistent with the units on the right side?
(a) (d) (b) (e)
(c) (f)
4. In the equation y
cn at 2 you wish to determine the integer value (1, 2, etc.) of the ex-
ponent n . The dimensions of y, a,
and t
are known. It is also known that c
has no dimen-
sions. Can dimensional analysis be used to determine n ? t B2x a v at v3 2ax2x vt 1 2at2 v at 1 2at3x vt
CHECK YOUR UNDERSTANDING
Scientists use mathematics to help them describe how the physical universe works, and trigonometry is an important branch of mathematics. Three trigonometric functions are utilized throughout this text. They are the sine, the cosine, and the tangent of the angle
TRIGONOMETRY
1.4
Problem-solving insight
You can check for errors that may have arisen during algebraic manipulations by performing a dimensional analysis on the final expression.
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1.4 TRIGONOMETRY 7
(Greek theta), abbreviated as sin , cos , and tan , respectively. These functions are de- fined below in terms of the symbols given along with the right triangle in Figure 1.5.
DEFINITION OF SIN , COS , AND TAN
(1.1)
(1.2)
(1.3)
h length of the hypotenuse of a right triangle ho length of the side opposite the angle ha length of the side adjacent to the angle
tan
ho ha
cos
ha h
sin
ho h
The sine, cosine, or tangent may be used in calculations such as that in Example 4, de- pending on which side of the triangle has a known value and which side is asked for. However, the choice of which side of the triangle to label ho (opposite) and which to label ha (adjacent) can be made only after the angle is identified. Often the values for two sides of the right triangle in Figure 1.5 are available, and the value of the angle is unknown. The concept of inverse trigonometric functions plays an important role in such situations. Equations 1.4-1.6 give the inverse sine, inverse cosine, and inverse tangent in terms of the symbols used in the drawing. For instance, Equation 1.4 is read as “ equals the angle whose sine is ho/h.”
(1.4)
(1.5)
(1.6)
tan1 ho ha
cos1 ha h
sin1 ho h
The sine, cosine, and tangent of an angle are numbers without units, because each is the ratio of the lengths of two sides of a right triangle. Example 4 illustrates a typical applica- tion of Equation 1.3.
On a sunny day, a tall building casts a shadow that is 67.2 m long. The angle between the sun’s rays and the ground is 50.0°, as Figure 1.6 shows. Determine the height of the building.
Reasoning We want to find the height of the building. Therefore, we begin with the colored right triangle in Figure 1.6 and identify the height as the length ho of the side opposite the an- gle . The length of the shadow is the length ha of the side that is adjacent to the angle . The ratio of the length of the opposite side to the length of the adjacent side is the tangent of the an- gle , which can be used to find the height of the building. Solution We use the tangent function in the following way, with 50.0 and ha 67.2 m:
(1.3)
The value of tan 50.0 is found by using a calculator.
80.0 m ho ha tan (67.2 m)(tan 50.0) (67.2 m)(1.19)
tan
ho ha
Example 4 Using Trigonometric Functions
90°
h = hypotenuse
θ
ho = length of side opposite the angle
ha = length of side adjacent to the angle θ
θ
= 50.0° θ
= 67.2 mh a
ho
Figure 1.5 A right triangle.
Figure 1.6 From a value for the angle and the length ha of the shadow, the height ho of the building can be found using trigonometry.
Problem-solving insight
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The use of 1 as an exponent in Equations 1.4-1.6 does not mean “take the reciprocal.” For instance, (ho/ha) does not equal 1/tan (ho/ha). Another way to express the in- verse trigonometric functions is to use arc sin, arc cos, and arc tan instead of and tan Example 5 illustrates the use of an inverse trigonometric function. 1 cos. 1, sin1, tan1
8 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS
A lakefront drops off gradually at an angle , as Figure 1.7 indicates. For safety reasons, it is necessary to know how deep the lake is at various distances from the shore. To provide some information about the depth, a lifeguard rows straight out from the shore a distance of 14.0 m and drops a weighted fishing line. By measuring the length of the line, the lifeguard determines the depth to be 2.25 m. (a) What is the value of ? (b) What would be the depth d of the lake at a distance of 22.0 m from the shore?
Reasoning Near the shore, the lengths of the opposite and adjacent sides of the right triangle in Figure 1.7 are ho 2.25 m and ha 14.0 m, relative to the angle . Having made this iden- tification, we can use the inverse tangent to find the angle in part (a). For part (b) the opposite and adjacent sides farther from the shore become ho d and ha 22.0 m. With the value for obtained in part (a), the tangent function can be used to find the unknown depth. Considering the way in which the lake bottom drops off in Figure 1.7, we expect the unknown depth to be greater than 2.25 m.
Solution (a) Using the inverse tangent given in Equation 1.6, we find that
(b) With 9.13°, the tangent function given in Equation 1.3 can be used to find the unknown depth farther from the shore, where ho d and ha 22.0 m. Since tan ho/ha, it follows that
which is greater than 2.25 m, as expected.
3.54 m d (22.0 m)(tan 9.13) ho ha tan
9.13
tan1 ho ha tan1 2.25 m 14.0 m
Example 5 Using Inverse Trigonometric Functions
14.0 m
2.25 m
d
22.0 m
θ
Figure 1.7 If the distance from the shore and the depth of the water at any one point are known, the angle can be found with the aid of trigonometry. Knowing the value of is useful, be- cause then the depth d at another point can be determined.
The right triangle in Figure 1.5 provides the basis for defining the various trigonometric functions according to Equations 1.1-1.3. These functions always involve an angle and two sides of the triangle. There is also a relationship among the lengths of the three sides of a right triangle. This relationship is known as the Pythagorean theorem and is used often in this text.
PYTHAGOREAN THEOREM
The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides:
(1.7)h 2 ho 2 ha 2
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1.5 SCALARS AND VECTORS 9
The volume of water in a swimming pool might be 50 cubic meters, or the winning time of a race could be 11.3 seconds. In cases like these, only the size of the numbers mat- ters. In other words, how much volume or time is there? The 50 specifies the amount of water in units of cubic meters, while the 11.3 specifies the amount of time in seconds. Volume and time are examples of scalar quantities. A scalar quantity is one that can be described with a single number (including any units) giving its size or magnitude. Some other common scalars are temperature (e.g., 20 C) and mass (e.g., 85 kg). While many quantities in physics are scalars, there are also many that are not, and for these quantities the magnitude tells only part of the story. Consider Figure 1.8, which de- picts a car that has moved 2 km along a straight line from start to finish. When describing the motion, it is incomplete to say that “the car moved a distance of 2 km.” This statement would indicate only that the car ends up somewhere on a circle whose center is at the start- ing point and whose radius is 2 km. A complete description must include the direction along with the distance, as in the statement “the car moved a distance of 2 km in a direc- tion 30 north of east.” A quantity that deals inherently with both magnitude and direction is called a vector quantity. Because direction is an important characteristic of vectors, ar- rows are used to represent them; the direction of the arrow gives the direction of the vec- tor. The colored arrow in Figure 1.8, for example, is called the displacement vector, because it shows how the car is displaced from its starting point. Chapter 2 discusses this particular vector. The length of the arrow in Figure 1.8 represents the magnitude of the displacement vector. If the car had moved 4 km instead of 2 km from the starting point, the arrow would have been drawn twice as long. By convention, the length of a vector arrow is propor- tional to the magnitude of the vector. In physics there are many important kinds of vectors, and the practice of using the length of an arrow to represent the magnitude of a vector applies to each of them. All forces, for instance, are vectors. In common usage a force is a push or a pull, and the di- rection in which a force acts is just as important as the strength or magnitude of the force. The magnitude of a force is measured in SI units called newtons (N). An arrow representing a force of 20 newtons is drawn twice as long as one representing a force of 10 newtons. The fundamental distinction between scalars and vectors is the characteristic of direc- tion. Vectors have it, and scalars do not. Conceptual Example 6 helps to clarify this dis- tinction and explains what is meant by the “direction” of a vector.
SCALARS AND VECTORS
1.5
N
S
WE
30.0°
Start
Finish
2 km
Figure 1.8 A vector quantity has a magnitude and a direction. The colored arrow in this drawing represents a dis- placement vector.
There are places where the temperature is 20 C at one time of the year and 20 C at an- other time. Do the plus and minus signs that signify positive and negative temperatures imply that temperature is a vector quantity? (a) Yes (b) No
Reasoning A hallmark of a vector is that there is both a magnitude and a physical direction associated with it, such as 20 meters due east or 20 meters due west.
Answer (a) is incorrect. The plus and minus signs associated with 20 C and 20C do not convey a physical direction, such as due east or due west. Therefore, temperature cannot be a vector quantity.
Answer (b) is correct. On a thermometer, the algebraic signs simply mean that the tempera- ture is a number less than or greater than zero on the temperature scale being used and have nothing to do with east, west, or any other physical direction. Temperature, then, is not a vec- tor. It is a scalar, and scalars can sometimes be negative.
Conceptual Example 6 Vectors, Scalars, and the Role of Plus and Minus Signs
Often, for the sake of convenience, quantities such as volume, time, displacement, velocity, and force are represented in physics by symbols. In this text, we write vectors in
Each of these racers in the Tour de France has a velocity as they approach the finish line on the Champs Élysées in Paris. Velocity is an example of a vector quantity, because it has a magnitude (the speed of the racer) and a direction (toward the finish line). (Owen Franken/Corbis Images)
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boldface symbols (this is boldface) with arrows above them* and write scalars in italic symbols (this is italic). Thus, a displacement vector is written as “ , due east,” where the is a boldface symbol. By itself, however, separated from the direction, the mag- nitude of this vector is a scalar quantity. Therefore, the magnitude is written as “A750 m,” where the A is an italic symbol without an arrow. A B A B 750 m
10 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS
*Vectors are also sometimes written in other texts as boldface symbols without arrows above them.
ADDITION
Often it is necessary to add one vector to another, and the process of addition must take into account both the magnitude and the direction of the vectors. The simplest situa- tion occurs when the vectors point along the same direction-that is, when they are colin- ear, as in Figure 1.9. Here, a car first moves along a straight line, with a displacement vector of 275 m, due east. Then, the car moves again in the same direction, with a displacement vector of 125 m, due east. These two vectors add to give the total dis- placement vector , which would apply if the car had moved from start to finish in one step. The symbol is used because the total vector is often called the resultant vector. With the tail of the second arrow located at the head of the first arrow, the two lengths sim- ply add to give the length of the total displacement. This kind of vector addition is identi- cal to the familiar addition of two scalar numbers (2 3 5) and can be carried out here only because the vectors point along the same direction. In such cases we add the individ- ual magnitudes to get the magnitude of the total, knowing in advance what the direction must be. Formally, the addition is written as follows:
Perpendicular vectors are frequently encountered, and Figure 1.10 indicates how they can be added. This figure applies to a car that first travels with a displacement vec- tor of 275 m, due east, and then with a displacement vector of 125 m, due north. The two vectors add to give a resultant displacement vector . Once again, the vectors to be added are arranged in a tail-to-head fashion, and the resultant vector points from the tail of the first to the head of the last vector added. The resultant displacement is given by the vector equation
The addition in this equation cannot be carried out by writing R 275 m 125 m, be- cause the vectors have different directions. Instead, we take advantage of the fact that the triangle in Figure 1.10 is a right triangle and use the Pythagorean theorem (Equation 1.7). According to this theorem, the magnitude of is
The angle in Figure 1.10 gives the direction of the resultant vector. Since the lengths of all three sides of the right triangle are now known, either sin , cos , or tan can be used R v(275 m)2 (125 m)2 302 m R B
R B
A B
B B
R B
B B
A B
R B
275 m, due east 125 m, due east 400 m, due east
R B
A B
B B
R B R B
B B
A B
VECTOR ADDITION AND SUBTRACTION
1.6
( The answer is given at the end of the book. )
5. Which of the following statements, if any, involves a vector? (a) I walked 2 miles along the beach. (b) I walked 2 miles due north along the beach. (c) I jumped off a cliff and hit the water traveling at 17 miles per hour. (d) I jumped off a cliff and hit the water traveling straight down at a speed of 17 miles per hour. (e) My bank account shows a negative balance of 25 dollars. CHECK YOUR UNDERSTANDING
N
S
WE
Tail-to-head
Start Finish A
R
B
N
S
WE
Tail-to-head
Start
R
B
Finish
90°
θ
A
Figure 1.9 Two colinear displacement vectors and add to give the resultant displacement vector . R B B B A B
Figure 1.10 The addition of two per- pendicular displacement vectors and gives the resultant vector . R B B B A B
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1.6 VECTOR ADDITION AND SUBTRACTION 11
to determine . Noting that tan B/A and using the inverse trigonometric function, we find that:
Thus, the resultant displacement of the car has a magnitude of 302 m and points north of east at an angle of 24.4. This displacement would bring the car from the start to the finish in Figure 1.10 in a single straight-line step. When two vectors to be added are not perpendicular, the tail-to-head arrangement does not lead to a right triangle, and the Pythagorean theorem cannot be used. Figure 1.11a illustrates such a case for a car that moves with a displacement of 275 m, due east, and then with a displacement of 125 m, in a direction 55.0° north of west. As usual, the re- sultant displacement vector is directed from the tail of the first to the head of the last vector added. The vector addition is still given according to
However, the magnitude of is not RA B, because the vectors and do not have the same direction, and neither is it , because the vectors are not perpendicular, so the Pythagorean theorem does not apply. Some other means must be used to find the magnitude and direction of the resultant vector. One approach uses a graphical technique. In this method, a diagram is constructed in which the arrows are drawn tail to head. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor, perhaps). Then, the length of the arrow representing the resultant vector is measured with a ruler. This length is con- verted to the magnitude of the resultant vector by using the scale factor with which the drawing is constructed. In Figure 1.11b, for example, a scale of one centimeter of arrow length for each 10.0 m of displacement is used, and it can be seen that the length of the ar- row representing is 22.8 cm. Since each centimeter corresponds to 10.0 m of displace- ment, the magnitude of is 228 m. The angle , which gives the direction of , can be measured with a protractor to be 26.7° north of east. R B R B R B R 2A2 B2 B B A B R B R B A B B B R B B B A B
tan1 B A tan1 125 m 275 m 24.4
SUBTRACTION
The subtraction of one vector from another is carried out in a way that depends on the following fact. When a vector is multiplied by 1, the magnitude of the vector remains the same, but the direction of the vector is reversed. Conceptual Example 7 illustrates the meaning of this statement.
Finish
Start
Tail-to-head
55.0°
55.0°
( ) a
( ) b
22.8 cm 4 8 12 200 24 cm 16 28
N
S
WE
θ
θ
A
R
B
A
R B
Figure 1.11 (a) The two displacement vectors and are neither colinear nor perpendicular but even so they add to give the resultant vector . (b) In one method for adding them together, a graphical technique is used. R B B B A B
Consider two vectors described as follows:
1. A woman climbs 1.2 m up a ladder, so that her displacement vector is 1.2 m, upward along the ladder, as in Figure 1.12a.
2. A man is pushing with 450 N of force on his stalled car, trying to move it eastward. The force vector that he applies to the car is 450 N, due east, as in Figure 1.13a.
What are the physical meanings of the vectors and ?
(a) points upward along the ladder and has a magnitude of 1.2 m; points due east and has a magnitude of 450 N. (b) points downward along the ladder and has a magnitude of 1.2 m; points due west and has a magnitude of 450 N. (c) points downward along the ladder and has a magnitude of 1.2 m; points due west and has a magnitude of 450 N.
Reasoning A displacement vector of is (1) . The presence of the (1) factor reverses the direction of the vector, but does not change its magnitude. Similarly, a force vector of has the same magnitude as the vector but has the opposite direction.
Answer (a) and (b) are incorrect. While scalars can sometimes be negative, magnitudes of vectors are never negative.
F B
F B
D B
D B F B
D B
F B
D B
F B
D B
F B
D B
F B
D B Conceptual Example 7 Multiplying a Vector by 1 (a)
D
(b)
-D
Figure 1.12 (a) The displacement vec- tor for a woman climbing 1.2 m up a ladder is . (b) The displacement vec- tor for a woman climbing 1.2 m down a ladder is . D B D B
2762T_ch01_001-027.qxd 4/14/08 10:12 PM Page 11
Answer (c) is correct. The vectors and have the same magnitudes as and , but point in the opposite directions, as indicated in Figures 1.12b and 1.13b.
Related Homework: Problem 33
F B
D B
F B
D B
In practice, vector subtraction is carried out exactly like vector addition, except that one of the vectors added is multiplied by a scalar factor of 1. To see why, look at the two vectors and in Figure 1.14a. These vectors add together to give a third vector , according to . Therefore, we can calculate vector as , which is an example of vector subtraction. However, we can also write this result as ( ) and treat it as vector addition. Figure 1.14b shows how to calculate vec- tor by adding the vectors and . Notice that vectors and are arranged tail to head and that any suitable method of vector addition can be employed to determine . A B B B C B B B C B A B B B C B A B B B C B A B A B B B A B C B C B B B A B
( The answers are given at the end of the book .) 6. Two vectors and are added together to give a resultant vector : . The magnitudes of and are 3 m and 8 m, respectively, but the vectors can have any orientation. What is (a) the maximum possible value and (b) the minimum possible value for the magnitude of ?
7. Can two nonzero perpendicular vectors be added together so their sum is zero?
8. Can three or more vectors with unequal magnitudes be added together so their sum is zero?
9. In preparation for this question, review Conceptual Example 7. Vectors and satisfy the vector equation 0. (a) How does the magnitude of compare with the magnitude of ? (b) How does the direction of compare with the direction of ? 10. Vectors , , and satisfy the vector equation , and their magnitudes are related by the scalar equation A 2 B 2 C 2. How is vector oriented with respect to vector ? 11. Vectors , , and satisfy the vector equation , and their magnitudes are related by the scalar equation A B C . How is vector oriented with respect to vector ? B B A B C B B B A B C B B B A B B B A B C B B B A B C B B B A B A B B B A B B B B B A B B B A B
R B
B B
A B
B B
A B
R B
R B
B B
A B CHECK YOUR UNDERSTANDING
A
B
C = A + B
( ) a
B-
C
Tail-to-head
( ) b
A = C - B
Figure 1.14 (a) Vector addition according to . (b) Vector subtraction according to ( ). B B C B B B C B A B B B A B C B
VECTOR COMPONENTS
Suppose a car moves along a straight line from start to finish, as in Figure 1.15, the corresponding displacement vector being . The magnitude and direction of the vector give the distance and direction traveled along the straight line. However, the car could also arrive at the finish point by first moving due east, turning through 90°, and then moving due north. This alternative path is shown in the drawing and is associated with the two dis- placement vectors and . The vectors and are called the x vector component and the y vector component of . Vector components are very important in physics and have two basic features that are apparent in Figure 1.15. One is that the components add together to equal the original vector:
The components and , when added vectorially, convey exactly the same meaning as does the original vector : they indicate how the finish point is displaced relative to the starting point. The other feature of vector components that is apparent in Figure 1.15 is that and are not just any two vectors that add together to give the original vector : they are perpen- dicular vectors. This perpendicularity is a valuable characteristic, as we will soon see. Any type of vector may be expressed in terms of its components, in a way similar to that illustrated for the displacement vector in Figure 1.15. Figure 1.16 shows an arbitrary r B y Bx B r B y Bx B
y Bx Br B
r B
y Bx By Bx B
r Br B
THE COMPONENTS OF A VECTOR
1.7
Figure 1.13 (a) The force vector for a man pushing on a car with 450 N of force in a direction due east is . (b) The force vector for a man pushing on a car with 450 N of force in a direc- tion due west is . F B F B
(a)
(b)
F
-F B
B
Start
Finish
90°
N
S
WE
x
yr
Figure 1.15 The displacement vector and its vector components and . y Bx B
r B
12 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS
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1.7 THE COMPONENTS OF A VECTOR 13
DEFINITION OF VECTOR COMPONENTS
In two dimensions, the vector components of a vector are two perpendicular vectors x and y that are parallel to the x and y axes, respectively, and add together vectori- ally so that xy .A B A B A B A B A B A B
In general, the components of any vector can be used in place of the vector itself in any calculation where it is convenient to do so (Problem-solving insight). The values calcu- lated for vector components depend on the orientation of the vector relative to the axes used as a reference. Figure 1.18 illustrates this fact for a vector by showing two sets of axes, one set being rotated clockwise relative to the other. With respect to the black axes, vector has perpendicular vector components x and y; with respect to the colored ro- A B A B A B A B tated axes, vector has different vector components x and y . The choice of which set of components to use is purely a matter of convenience. A B A B A B
Ay
+y
+x
Ax
A
θ
+y
+x
+y´
+x´
Ax
A´y
A´x
A
Ay
SCALAR COMPONENTS
It is often easier to work with the scalar components, Ax and Ay (note the italic symbols), rather than the vector components x and y. Scalar components are positive or negative numbers (with units) that are defined as follows: The scalar component Ax has a magnitude equal to that of x and is given a positive sign if x points along the axis and a negative sign if x points along the axis. The scalar component Ay is defined in a similar manner. The following table shows an example of vector and scalar components:
Vector Components Scalar Components Unit Vectors
x 8 meters, directed along the x axis meters x (8 meters) y 10 meters, directed along the y axis meters y (10 meters)
In this text, when we use the term “component,” we will be referring to a scalar compo- nent, unless otherwise indicated. Another method of expressing vector components is to use unit vectors. A unit vector is a vector that has a magnitude of 1, but no dimensions. We will use a caret (^) to distin- guish it from other vectors. Thus,
is a dimensionless unit vector of length l that points in the positive x direction, and is a dimensionless unit vector of length l that points in the positive y direction.
These unit vectors are illustrated in Figure 1.19. With the aid of unit vectors, the vector components of an arbitrary vector can be written as Ax and Ay , where Ax and Ay are its scalar components (see the drawing and the third column of the table above). The vector is then written as Ax Ay .
congrats on your 1000th :)
Congrats cyber u da man can't believe u made 1000 happy 10000 :)
Contest
1st: Tachyon Galaxy Box
2nd: 15 Tachyon Galaxy Packs
3rd: 10 Tachyon Galaxy Packs
4th: 5 Tachyon Galaxy Packs
5th: 2 Tachyon Galaxy Packs
and congratulations on 1000 videos keep up the great work
can you PLLLLEEEEASEEEE do these requests?
1: After tachyon galaxy comes out, update galaxy deck?
2: Update the heraldic beasts deck
3: make a heroic challenger deck or until the number hunters pack comes out with the new heroic cards
4: and in time, when the other packs come out can you make the barian emperors' deck profiles
love your vids cyber, keep up the great work =D
It's funny that you mentioned the meta and X-Sabers; I'm actually working on making X-Sabers meta again. I think they're probably around tier 1.5 with the build I'm currently using.
1000! Congrats!
Congrats long time sub keep it up
You should do more of these, they are so awesome!
i created my own playmat its an e-hero one there my fave do u know how i can make it like a real mat
Hey man howd you get these packs? i mean like i was checking strikezone and they dont have any packs in stock?
Thank you for uploading such great videos man.
thank you :D
I would be interested, how does your apartment looks like xD
Wow 1000 video congrats that's cool .
awesome i love your videos your my favorite yugituber
Thank You Cyberknight for giving us an epic 1000th video
where do you get all these booster packs and how much have you payed for them?
i barely like video but this and everyone of your videos are alsome XD, i watch videos when i am doing homework so i like around 15-50 min videos i agree with cyber knight
When Caleb drinks a red Bull before an opening 😂
How many megalosaurus do you need for a mermail deck?
congratulations to ypour 1000th video!!!
Wow.... I just noticed, this video was uploaded on my birthday!!! This would have been an awesome present. I love watching unboxings and want to start one myself!!! I need a tripod though :/
Even that I dont like yugioh i really enjoyed your video! :)
omg this is epic so much packs :D
how much do you paid for that
yes im looking for madolches got any still?
Keep up the videos ur amazing man!!!
How much did all these cost in total I would like to know
+Extreme Dreamz it would be like 200-300 $ XD
You buy all these packs separately or do they do a bundle deal?
Very Much holos on you TSHD booster pack!! amazing! i like it!
Congrats man!!!!!!!!!!!!!!!!!!!!!!!!!
Hey guys, im kinda yugioh fan and i want to start collecting cards (and also getting my dragon deck).. where do i start ?
Which booster box should i buy if i love wind-ups?
What do you do with the cards you don't use?
AWSOME OPENING!!!!!!!!!!!!!!!! Eyes are a little dry, but its very mush so worth it! Next time maybe 150?
Love watching ur videos!!!!!!!
lol i love these videos cybernight
Cyberknight8610, you are amazing. How did you get all these packs?
for a contest made a box of return of the duelist? or a special edition box?
New drinking game:
Every time CyberKnight mentions slushy take a shot.
Hey cyberknight8610 what do you want for the gateway of six sam?
What do u want for the mermail cards
how about an entire deck plus extra deck for the contest and congrats on reaching 1000 videoes
And now you're 4000 subs off 100,000! Congrats.
congrats! nice holos!
No recap?
When the Blue-Eyed Structure deck comes out for Japan, could you do an opening of it?
Did you end up selling your last rare holos to fund this pack opening I'm just curious.
Best yugioh channel ever
HOW CAN THE DARKNESS BE SHINNING!?!?!, WHAT SORCERY IS THIS, Mind = Blown
How much are you selling for a play set of Madolche magileine.???
Everyone donate packs to him so he can do a 100,000 pack opening when he reaches 100,000 subs!!!
+TheBlacksmith21 will do XD
Cyberknight8610 Chaos King Archfiend was also one of the cards that was the result of a drawing competition :)
Congrats dude
Cyberknight i have a question concerning what sites to order my english booster boxes, specifically the new Judgment of the light thats comming out soon. can you please pm me or something to tell me what sites you order your booster boxes from?