I have a hard time understanding physics when there's no too much theory, but your videos are the only ones who make me understand 100% everything in physics. I want to say that thanks to your videos I got the best score in physics of 100%
@@stop2759 Sorry I was a bit rude; I was just trying to sound blunt to give this guy a reality check. If you can't understand the basic maths, then you're going to have trouble with all physics in later years.
@@pubgplayer1720 physics needs to be explained with theory not just the use of equations, as that will help students broaden or lighten the topic they're dealing with.
For the past five decades, having flunked out of premed (remember, this was the '60s) I have still yearned to truly understand physics, chemistry, et al., simply for the sheer pleasure of understanding how the Universe works, what it is made of and its history. I've tried periodically many times over the years to initiate this understanding, since I have a large quantity of science textbooks I've purchased over the years (some applying calculus and some not). I start out full of resolve and optimism, I solve some of the problems. But soon enough I'm not able to solve them anymore and my energy (I'm using the term strictly in its colloquial metaphoric sense-- I understand its definition in physics) flags and I give up. I was always left with a feeling of frustration, because the yearning remained, despite the futility. Then, I discovered your tutorials and now I'm finally grokking (again with the '60s) these concepts and solving the problems. .And, at 75 years old, just in time. Grazia tanto, Signor Professore!
I love how simple and easy you make these concepts seem, my tutors over complicate EVERYTHING and their lectures leave me more confused than before I started! A class mate recommended your videos over the lectures and I'm so glad they did because I'm now understanding things in 5 minutes that before would of taken me weeks of sifting through my tutors overcomplicated jargon to reach the same understanding. I wish all professors were as clear cut and concise as you, keep up the good work educating the masses, love your content!
So grateful that these videos exist! Unfortunately, professors can have a great understanding of physics but little ability to pass that understanding on. You, however, do have that ability!
Thank you so much, I have been trying to figure this out for 2 days. The teachers lecture is very unhelpful in using the formula even though he took 2-3 hours to explain it, the textbook literally has only solutions for F=kx which made it impossible to even use the textbook. I can't believe how hard it was to even find a proper explanation on google, finding your videos is a godsend. I can finally answer the 4 questions the teacher has posted. I really really like physics and the idea but good explanations are so far and few.
eljko Milković, an academic and inventor from Novi Sad, has done something great that has not been done by any Serbian inventor before. Milković invention of the mechanical oscillator is widely used worldwide, a testament to the fact that over 500 foreign companies use, sell and manufacture pendulum-based machines used in the heavy industry. The purpose of the two-stage meganic oscillator is multifaceted, because the character of the machine (two-arm lever with pendulum) allows its use as a press, water pumps, compressor, crusher, power generator, mini power plants.
Professor, I had a question. If you would answer this, I will be very grateful.The question is: An object is in free fall when the force acting on it is exclusively gravitational. But why then the moon is in free fall? Isn't there centrifugal force acting on it? Thanks professor. Lots of love.
The moon is in free fall around the Earth. However, due to the fact that the moon has sufficient tangential speed, the moon is continuing to overshoot its target, which is why it doesn't crash in to the Earth. If the orbit were a perfect circle, the force of gravity would directly match the net force needed to cause centripetal acceleration. The force of gravity would be entirely perpendicular to the moon's velocity, which means there is zero work done by the force of gravity on the moon, and zero change in speed. The orbit is not a perfect circle, so to a lesser extent, the moon does speed up slightly and slow down slightly. The path of the orbit is elliptical, with the Earth-Moon center of mass (which is inside of Earth) at one of the focus points. You can look up the apogee (farthest distance from Earth) and perigee (closest distance from Earth) of the moon to see how far it varies from a perfect circle. There is the parameter called eccentricity that is also a measure of how non-circular an orbital shape is. The moon does speed up as it approaches perigee and slow down as it approaches apogee, so gravity does do a little bit of work on the moon during this path, but not enough to put it on a crash trajectory with Earth.
Centrifugal force from motion of object balances the gravitational potential energy so moon doesn't moves away from the orbit and continues constant circular motion
thanks! this video is really helpful especially during this time of pandemic where im not attending school. honestly this is better explained than my teacher! Keep up the good videos!! :)
If a ball is on the ground,its potential energy is 0, but if I dig a circle around the ball so that the ball looks like as if it's on top of a tower, it suddenly has potential energy. PLEASE EXPLAIN HOW Prof.
ooh good one! i think it would still have zero potential energy, it would only have potential energy if it were rolled off the edge such that it could fall. that may or may not be right!
Gravitational potential energy is relative to whatever you choose to be your datum. If you calculate total mechanical energy based on that datum, it will be conserved in a closed system either way. You just can't change the datum at any point during your evaluation.
Good question. It is arbitrary where we define GPE to equal zero. A ball sitting on the ground doesn't necessarily have truly zero GPE, we just define it to have zero GPE by convention. It is mathematically impractical to calculate how much GPE it truly has. We define its initial GPE to equal zero. We then dig a hole 1 meter deep. We let the 1kg ball roll toward the hole. The ball falls to the bottom of the hole, and its GPE becomes -9.8 Joules. The ball lost 9.8 Joules of GPE that became thermal energy, when it falls and comes to rest. When using the universal law of gravitation, we set GPE = 0 at r=infinity, but this is just for mathematical convenience. When we do this, all GPE's end up becoming negative. It is a simple consequence of picking r=infinity as our datum for GPE for mathematical convenience. There is no such thing as absolute GPE, the way we calculate it in Physics.
If physics is the study of the natural physical world, mathematically observed, noted, studied, & expressed? What about the vibration behind the thermal energy output by our first, most important muscle, the brain, teacher?
good question, i think when it's g and not a, we just use the scalar value rather than the vector so we ignore direction. may want to double check but i think that's correct.
Because acceleration due to gravity is constant and points in the negative direction, acting on all projectiles and decreasing their vertical velocity all the time.
How is the energy of a closed system conserved if ,according to the 1st law of thermodynamics, says that adding heat can cause a change in internal energy? Even though the system has no mass transfer the internal energy can still change. Maybe I just don’t have the right idea
Energy can be transferred in or out of a system. That's not a violation of the 1st law. The energy has to come from somewhere or go somewhere. It's conserved.
Is the conservation of the energy found in the way it is transformed? I'm trying to understand what you mean by conservation of energy...like it's saved? or more precisely not lost? because it goes from one form to another? Can the conservation be measured? Thank you for your videos, they're great ☺.
it means the total amount of energy remains constant for a closed system! the amount of energy can be measured, yes, we just have to account for all the different forms of energy. thanks for watching!
Conservation of energy means you cannot create or destroy energy. It only changes form, and changes what object contains it. The sum total energy in an isolated system will equal a constant value. Isolated system means no mass or energy crosses the boundaries of the system.
Just curious. In the case of radioactive decay, let say Uranium 235 has a half life of 200 years. So, for a 10 KG Uranium 235, 200 years later, it will reduce to 5KG. Then where is the energy of another 5KG? If we were to work backward, if the earth is more than 10 billions years old. 10 billions divided by 200 years. That mean a 10 KG Uranium 235 around 10 billions years ago, its size before the helf life can be so huge right? Then, what has happened to the energy? If we use the Sun which is doing fusion every moment. The energy should be transformed into heat energy. then during the travel to the earth, where is the lost energy?
" Then where is the energy of another 5KG?" The 5kg doesn't cease to exist, it just changes form. When 10 kg of U235 decay, 5 kg become something else, and 5 kg are still remaining as the original isotope of U235. There are biproducts when U-235 decays, and in its natural decay cycle, the next atom is Thorium-231. That means only about 4 amu of mass are annihilated and scattered as energy, while the rest remains as matter in the biproduct of Thorium. The natural decay chain of U235 eventually settles on lead-207, which is stable. When U235 decays to Pb207, there are 28 amu of mass that get emitted as gamma rays, alpha particles, and free electrons, the gamma rays of which get absorbed and converted to thermal energy. The alpha particles become natural sources of helium on our planet. If you look at the energy balance of Earth, based on the incoming solar radiation, outgoing thermal radiation in the form of infrared, you find that you need to add energy generated in the Earth's interior to make up the balance. This energy comes from the radioactive decay of naturally existing radioactive material
"If we use the Sun which is doing fusion every moment. The energy should be transformed into heat energy. then during the travel to the earth, where is the lost energy? " The energy is conserved on its journey from the sun to the Earth. It is just that most of it, continues to travel to the other planets and to deep space. Only a small fraction based on the angular size that the Earth would be in a hypothetical sky from the sun, actually makes it to Earth. But everything emitted in that angular fraction of the sun's surface, does indeed make it to Earth. You can calculate the extraterrestrial Watts/meter^2 that you expect at Earth's orbit, based on the inverse square law, and comparing it to measured data, it is indeed consistent with the sun's total luminosity and rate of nuclear fusion in its core.
If we are energy then once our bodies (matter) die, will our energy continue on? Since energy can not be destroyed then this kind of makes sense. This of course doesn't mean that energy of yours would have your consciousness nor memories.
Well as you say, we are energy in the sense that we are made of matter, and when we die, that matter persists. Our bodies decompose and the atoms go on to be part of other things, like the ground, and a tree, and the air, and so forth.
hi! thanks for the video! a great one...one note though - we do not need to know the mass on the comprehension section to calculate the speed ta 20 meters as it is cancelled out for the equation
Why it's not Noticed that Gravity Violets Energy of conservation ? How ? suppose there is lot's of object in space with zero energy ( forgot E= mc2 for this case ) then also imagine there Planet which try to To pull that all object towards it so object will gain kinetic energy and after impact it will release as heat think it planet's gravity have not spent anything to give that all energy to that object instead it gained some energy because of that added mass so next time it will so that same with more energy . Other hand think about black hole It always eats and becomes stronger and stronger over time just because of this strange nature of gravity in which it doesn't lose anything by doing any work ( work of attraction )
Your pretense that it started with zero energy when it was infinitely far away from the planet, is incorrect. We assign infinitely far away to have zero potential energy, but that is nothing more than a mathematical convenience. We don't know where true zero gravitational potential energy really would occur, and it is impractical to calculate, so we pick infinitely far away because that is what makes the math as simple as can be. It comes from integrating G*M*m/r^2 relative to r, which produces a +C arbitrary constant of integration. Since most of the time, we are only interested in differences in potential energy, rather than absolute potential energy (wich doesn't really exist), we keep it simple and assign C=0 when defining potential energy as it derives from N's law of gravitation.
@@carultch Nice Logical Answer but .. Both Situation ( Whether object are inside of the planet's finite Gravity field or Field with Infinite range from highest to lowest strength . ) Seems Violate the Conservation . Take Hypothetical example for Gravity as finite strength - Suppose Sun has Gravitational Field of 10AU unit and its very sharp boundary of it's G - field and there is an object with 1 kg mass , just meter away from sun's G- Field and it just need very little energy to push that object in side that sun's Gravity region ( 10AD assumed ) but because of Sun's gravitational acceleration that little triggering energy will end up millions of times stronger when that 1KG object will hit the sun . It seem Funny thought experiment but was just for saying - it doesn't matter if Gravitational field slowly fade out over distance or never ends towards infinity .
@@omsingharjit There are planets that are farther than 10 AU from the sun in the solar system, so your premise is already false. 10 AU is about the orbital position of Saturn. You have to go far beyond the edge of our galaxy to find a hard-stop to the sun's gravitational field. It does exist eventually, just not at the scale of 10 AU. Eventually, the expansion of the universe gets in the way, and causes space to be created between us and distant galaxies. Where enough space is being created, the gravitational waves cannot propagate fast enough to get to the distant galaxies, and such distant galaxies are no longer in gravitational contact with our own sun and our own galaxy. It is lucky that we are alive at a time when we can discover this, because other than the select few local galaxies like Andromeda and the Magellanic Clouds, most galaxies are far enough away from us that they are out of gravitational contact, and will perpetually move away. Eventually, they will move farther away than the observable universe, and no receding galaxies will be visible to us as evidence of the universal law of gravitation. How we account for the energy of universe expansion is still a mystery, and is a concept that cosmologists call dark energy.
@@hajimajinja4941 Because energy is conserved. There is no energy added to the swinging pendulum, when it fell from the person's face, to the opposite end of the room, and back to the person's face. The pendulum can go no higher, than the point at which it started. Time would need to run backwards, for the air drag and friction to do positive work on the swinging pendulum, which would speed it up. As crazy as this sounds, it is perfectly OK with the first law of thermodynamics. It is the second law of thermodynamics, that makes forces like friction and air drag slow objects down instead of speed them up.
Free fall pendulum is an oxymoron. The bob is still supported by the constraint force in the rod and pivot. There is no closed form equation for high amplitude pendulums, where you can't approximate sin(theta) as theta. You can solve it with an infinite series if you need to model it, but you cannot get an exact equation in closed form for the period. It also no longer undergoes simple harmonic motion, when you can no longer take credit for the approximation. Hyperphysics has a large amplitude pendulum period calculator, that enables you to see this effect.
if the object is falling, that means g is negative and h is negative . Right? so that's why you got positive answer (the negative cancelled themselves) .. instead of negative ???????? I always get confused with g. sometimes it be positive and sometimes it be negative. Please help!
The reason its sign changes, is that you assign a sign convention and coordinate system as you see fit to solve the problem. In any event, the gravitational field g acts downward, and your assignment of signs needs to account for this, as the downward direction fits your coordinate system. Sometimes it is more convenient to assign positive to the downward direction, other times it is more convenient to assign positive to the upward direction.
I am stuck on a problem, this is because there is no velocity given at ALL. How could you find the velocity without having the initial information (No time, or acceleration of the object falling was given)? I did try using the conservation of energy formula yet the answer came out to be no solution!
Using v² = u² + 2gh formula you can solve= it.... Here I am explaining.... v = ? , u = 40.5 m/s² , g = 9.8m/s² h = 20 m So, v² = (40.5)² + 2 × 9.8 × 20 = 2032.25 So, v = 45.08 m/s² So the final velocity is 45.08 m/s²
I have a hard time understanding physics when there's no too much theory, but your videos are the only ones who make me understand 100% everything in physics. I want to say that thanks to your videos I got the best score in physics of 100%
Yo this is literally primary school math drop out of high school if you don't understand any further...
Indian PUBG player Why so rude though?
@@stop2759 Sorry I was a bit rude; I was just trying to sound blunt to give this guy a reality check. If you can't understand the basic maths, then you're going to have trouble with all physics in later years.
@@pubgplayer1720 sometimes it's the teacher, the way it's explained, or it just isn't their passion. give it time and they'll be alright.
@@pubgplayer1720 physics needs to be explained with theory not just the use of equations, as that will help students broaden or lighten the topic they're dealing with.
For the past five decades, having flunked out of premed (remember, this was the '60s) I have still yearned to truly understand physics, chemistry, et al., simply for the sheer pleasure of understanding how the Universe works, what it is made of and its history. I've tried periodically many times over the years to initiate this understanding, since I have a large quantity of science textbooks I've purchased over the years (some applying calculus and some not). I start out full of resolve and optimism, I solve some of the problems. But soon enough I'm not able to solve them anymore and my energy (I'm using the term strictly in its colloquial metaphoric sense-- I understand its definition in physics) flags and I give up. I was always left with a feeling of frustration, because the yearning remained, despite the futility. Then, I discovered your tutorials and now I'm finally grokking (again with the '60s) these concepts and solving the problems. .And, at 75 years old, just in time. Grazia tanto, Signor Professore!
Wish you a happy and roaring life!
I wish all classes and lectures were just like this. Straight to the point and effecient
I love how simple and easy you make these concepts seem, my tutors over complicate EVERYTHING and their lectures leave me more confused than before I started! A class mate recommended your videos over the lectures and I'm so glad they did because I'm now understanding things in 5 minutes that before would of taken me weeks of sifting through my tutors overcomplicated jargon to reach the same understanding. I wish all professors were as clear cut and concise as you, keep up the good work educating the masses, love your content!
So grateful that these videos exist! Unfortunately, professors can have a great understanding of physics but little ability to pass that understanding on. You, however, do have that ability!
And, of course, you amply have the great understanding, inasmuch as you know a lot about the science stuff.
i can't express how satisfying it is to try to solve the problem at the end of the video and then get it right first try.
thanks prof dave!
Thank you so much, I have been trying to figure this out for 2 days. The teachers lecture is very unhelpful in using the formula even though he took 2-3 hours to explain it, the textbook literally has only solutions for F=kx which made it impossible to even use the textbook. I can't believe how hard it was to even find a proper explanation on google, finding your videos is a godsend. I can finally answer the 4 questions the teacher has posted. I really really like physics and the idea but good explanations are so far and few.
4:25 my heart is beating. im waiting for this moment! "did I calculate right?", "am i a loser?"! this is the most interesting moment
omg, I have the exact same feeling when I get the results of a physics test! Dang!
@@Matt-sc6gg same
eljko Milković, an academic and inventor from Novi Sad, has done something great that has not been done by any Serbian inventor before.
Milković invention of the mechanical oscillator is widely used worldwide, a testament to the fact that over 500 foreign companies use, sell and manufacture pendulum-based machines used in the heavy industry.
The purpose of the two-stage meganic oscillator is multifaceted, because the character of the machine (two-arm lever with pendulum) allows its use as a press, water pumps, compressor, crusher, power generator, mini power plants.
The intro felt like a fever dream ;-;
;_;
4:43 u can also do this with Vf^2 = Vi^2 = 2a(delta)x and get the same answer which proves these equations relates to one another :D
thanks Prof. Dave! This helped sooo much!!! All the love from the Philippines! :)))))))
Bakit konti lang subscribers niya??
Magaling kaya siyang mag turo
Dang who else teacher told us to watch this video to compete the assignment
You in college?
@@05bahaafarah11 Nope I was in 9th grade when I made this comment I’m a senior now
Hahahah yep
The only non indian guy I can search up when I am in despair.
hey its me again
love your simplicity !!
sir ur physics videos are absolutely clear with the concepts ..thus help students to understand them easily
Excelente, excelente, excelente, profesor Dave. ¡Saludos desde la República Dominicana!
Nice lecture sir... Love from India ....🇮🇳🇮🇳🇮🇳
you can also get the answer of the last question by using
v^2 = v0^2 + 2as formula
love the intro! iconic
Professor, I had a question. If you would answer this, I will be very grateful.The question is:
An object is in free fall when the force acting on it is exclusively gravitational. But why then the moon is in free fall? Isn't there centrifugal force acting on it?
Thanks professor.
Lots of love.
I guess the gravitational force is contributing to the required centrifugal force.Feel free to correct me if I am wrong. :)
@@adithyank3773 Centrifugal force does not exist.
Lynxica is correct. It would be centripetal force acting on the moon' gravitational pull.
The moon is in free fall around the Earth. However, due to the fact that the moon has sufficient tangential speed, the moon is continuing to overshoot its target, which is why it doesn't crash in to the Earth. If the orbit were a perfect circle, the force of gravity would directly match the net force needed to cause centripetal acceleration. The force of gravity would be entirely perpendicular to the moon's velocity, which means there is zero work done by the force of gravity on the moon, and zero change in speed.
The orbit is not a perfect circle, so to a lesser extent, the moon does speed up slightly and slow down slightly. The path of the orbit is elliptical, with the Earth-Moon center of mass (which is inside of Earth) at one of the focus points. You can look up the apogee (farthest distance from Earth) and perigee (closest distance from Earth) of the moon to see how far it varies from a perfect circle. There is the parameter called eccentricity that is also a measure of how non-circular an orbital shape is. The moon does speed up as it approaches perigee and slow down as it approaches apogee, so gravity does do a little bit of work on the moon during this path, but not enough to put it on a crash trajectory with Earth.
Centrifugal force from motion of object balances the gravitational potential energy so moon doesn't moves away from the orbit and continues constant circular motion
Your videos has helped me to understand physics
Thank you i really gained alot of value from this
I always watch till the end just to listen the end music :)
How can you find the maximum height of an object? I'm learning this unit in class and it was really confusing but your video explained it really well.
what an absolute saint. u are the best teacher!
He is the science god
thank u professor Dave I rlly needed this
Thank you PHYSICS JESUS!
How you'd come up with the final velocity of 1,640?
I know like how
Good teaching sir 👍💞
Sirr you explain this topic in very easy manner thank you......
The most useful video ever😍
This Man is really good
thanks! this video is really helpful especially during this time of pandemic where im not attending school. honestly this is better explained than my teacher! Keep up the good videos!! :)
lol dis is 2 years ago
@★SW4GGYP3NGU1N★ -OMG TYSM FOR 50 nah 10th or 11th maybe
Helpful & to the point..thank you!!
Love your videos ❤
Very helpful! Tysm for making this video :)
If a ball is on the ground,its potential energy is 0, but if I dig a circle around the ball so that the ball looks like as if it's on top of a tower, it suddenly has potential energy. PLEASE EXPLAIN HOW Prof.
ooh good one! i think it would still have zero potential energy, it would only have potential energy if it were rolled off the edge such that it could fall. that may or may not be right!
so if I am on top of a mountain and its highest point is at sea level. does that means I don't have PE until I get rolled off the mountain?
Gravitational potential energy is relative to whatever you choose to be your datum. If you calculate total mechanical energy based on that datum, it will be conserved in a closed system either way. You just can't change the datum at any point during your evaluation.
Was the potential created by removing the circle of dirt?
Good question. It is arbitrary where we define GPE to equal zero. A ball sitting on the ground doesn't necessarily have truly zero GPE, we just define it to have zero GPE by convention. It is mathematically impractical to calculate how much GPE it truly has.
We define its initial GPE to equal zero. We then dig a hole 1 meter deep. We let the 1kg ball roll toward the hole. The ball falls to the bottom of the hole, and its GPE becomes -9.8 Joules. The ball lost 9.8 Joules of GPE that became thermal energy, when it falls and comes to rest.
When using the universal law of gravitation, we set GPE = 0 at r=infinity, but this is just for mathematical convenience. When we do this, all GPE's end up becoming negative. It is a simple consequence of picking r=infinity as our datum for GPE for mathematical convenience. There is no such thing as absolute GPE, the way we calculate it in Physics.
If physics is the study of the natural physical world, mathematically observed, noted, studied, & expressed?
What about the vibration behind the thermal energy output by our first, most important muscle, the brain, teacher?
3:46 who else decided to slap the table
In the comprehension test, the object falls toward the ground, isn't the g = (-9.8m/s^2)? Why it is positive?
good question, i think when it's g and not a, we just use the scalar value rather than the vector so we ignore direction. may want to double check but i think that's correct.
@@ProfessorDaveExplains Thanks for the respond :)
Tysm 😭😭♥️♥️
Love from India ❤️
Can u tell why does the speed (v) decreases on going upwards?
Because acceleration due to gravity is constant and points in the negative direction, acting on all projectiles and decreasing their vertical velocity all the time.
U are a good guy professor Dave luv u sir♥️♥️
Thanks professor Dave, very informative. Pongan listo!
Thanks for the videos!
Very nice explanation.thank you
How did you get the initial velocity? Or the final velocity?
2:32 What about air resistance?
Yes
very good
when can i use g or -g? does velocity a vector?
Sir is both kinetic energy and potential energy are mechanical or thermal energy?
They both are considered mechanical energy.
Thermal energy ultimately is kinetic energy, except at the molecular level, instead of macroscopic.
Not sure how he got to v^2 = 1640. How did that work? I have a test on this sooooonnnn
all the arithmetic is there! don't forget to square that initial velocity.
@@ProfessorDaveExplains Could you please explain it briefly as a reply. I'm not able to understand how you got 1640. Would be appreciated.
Is it correlate with projectile motion. I use it and result is the same thanks
@@lorihunter1429 Thanks for that tip, I actually struggled with this problem at first but your comment made me finally solve the equation right.
Thank you very much
what are the potential energies of a Blue Shell vs Red Shell vs Green Shell? If you're in 1st vs 2nd vs 3rd place?
If you're a stud, green is all you need.
dude your intro song is meme lol it is pretty funny to and like your content of teaching others like a teacher but in a short way
Done.
I love you, you are one of the best teachers 😍
I did not understand what 4:44, well the working out meant 😔😖
what will be the mass of object at the start of swing, middle of swing and at the end of swing during one cycle.
How is the energy of a closed system conserved if ,according to the 1st law of thermodynamics, says that adding heat can cause a change in internal energy? Even though the system has no mass transfer the internal energy can still change. Maybe I just don’t have the right idea
Energy can be transferred in or out of a system. That's not a violation of the 1st law. The energy has to come from somewhere or go somewhere. It's conserved.
Professor Dave Explains I was not taking the surrounding into consideration. Thank you
how did you get 1,640 for Vf^2 ? I cant recognize the way of solution. All in all thumbs up for easy understanding
just simplify and solve! basic algebra, check out my algebra tutorials if it is confusing.
Thanks a lot ! 🤩
Please give a video about simple harmonic motion for pendulum with examples
great video! thanks!
he is the only reason im halfway passing school 💀
U r my god in physics and chemistry
He is the best teacher in those 2 subjects
So, which is correct? GPE > KE + SE + HE, or GPE > KE > SE + HE ?
This video helps me a lot thank you!💙💙💙💙
Your theme song should be on the grammys.
shut up kendal
Is the conservation of the energy found in the way it is transformed? I'm trying to understand what you mean by conservation of energy...like it's saved? or more precisely not lost? because it goes from one form to another? Can the conservation be measured? Thank you for your videos, they're great ☺.
it means the total amount of energy remains constant for a closed system! the amount of energy can be measured, yes, we just have to account for all the different forms of energy. thanks for watching!
Conservation of energy means you cannot create or destroy energy. It only changes form, and changes what object contains it. The sum total energy in an isolated system will equal a constant value. Isolated system means no mass or energy crosses the boundaries of the system.
how did you solve the problem in the checking comprehension?
Pls explain mechanical energy is conserved for a system here only ball is there
Just curious. In the case of radioactive decay, let say Uranium 235 has a half life of 200 years. So, for a 10 KG Uranium 235, 200 years later, it will reduce to 5KG. Then where is the energy of another 5KG? If we were to work backward, if the earth is more than 10 billions years old. 10 billions divided by 200 years. That mean a 10 KG Uranium 235 around 10 billions years ago, its size before the helf life can be so huge right? Then, what has happened to the energy? If we use the Sun which is doing fusion every moment. The energy should be transformed into heat energy. then during the travel to the earth, where is the lost energy?
" Then where is the energy of another 5KG?"
The 5kg doesn't cease to exist, it just changes form. When 10 kg of U235 decay, 5 kg become something else, and 5 kg are still remaining as the original isotope of U235. There are biproducts when U-235 decays, and in its natural decay cycle, the next atom is Thorium-231. That means only about 4 amu of mass are annihilated and scattered as energy, while the rest remains as matter in the biproduct of Thorium. The natural decay chain of U235 eventually settles on lead-207, which is stable.
When U235 decays to Pb207, there are 28 amu of mass that get emitted as gamma rays, alpha particles, and free electrons, the gamma rays of which get absorbed and converted to thermal energy. The alpha particles become natural sources of helium on our planet.
If you look at the energy balance of Earth, based on the incoming solar radiation, outgoing thermal radiation in the form of infrared, you find that you need to add energy generated in the Earth's interior to make up the balance. This energy comes from the radioactive decay of naturally existing radioactive material
"If we use the Sun which is doing fusion every moment. The energy should be transformed into heat energy. then during the travel to the earth, where is the lost energy?
"
The energy is conserved on its journey from the sun to the Earth. It is just that most of it, continues to travel to the other planets and to deep space. Only a small fraction based on the angular size that the Earth would be in a hypothetical sky from the sun, actually makes it to Earth. But everything emitted in that angular fraction of the sun's surface, does indeed make it to Earth.
You can calculate the extraterrestrial Watts/meter^2 that you expect at Earth's orbit, based on the inverse square law, and comparing it to measured data, it is indeed consistent with the sun's total luminosity and rate of nuclear fusion in its core.
If we are energy then once our bodies (matter) die, will our energy continue on? Since energy can not be destroyed then this kind of makes sense.
This of course doesn't mean that energy of yours would have your consciousness nor memories.
Well as you say, we are energy in the sense that we are made of matter, and when we die, that matter persists. Our bodies decompose and the atoms go on to be part of other things, like the ground, and a tree, and the air, and so forth.
@@dixztube Sure, I suppose that's one way of putting it.
Thankyou Sir!
hi! thanks for the video! a great one...one note though - we do not need to know the mass on the comprehension section to calculate the speed ta 20 meters as it is cancelled out for the equation
THANK YOU PHYSICS JESUS!!!!
I’m confused on how you got velocity
good video
Thanks
This is great
Why it's not Noticed that Gravity Violets Energy of conservation ?
How ?
suppose there is lot's of object in space with zero energy ( forgot E= mc2 for this case ) then also imagine there Planet which try to To pull that all object towards it so object will gain kinetic energy and after impact it will release as heat think it planet's gravity have not spent anything to give that all energy to that object instead it gained some energy because of that added mass so next time it will so that same with more energy .
Other hand think about black hole It always eats and becomes stronger and stronger over time just because of this strange nature of gravity in which it doesn't lose anything by doing any work ( work of attraction )
Your pretense that it started with zero energy when it was infinitely far away from the planet, is incorrect. We assign infinitely far away to have zero potential energy, but that is nothing more than a mathematical convenience. We don't know where true zero gravitational potential energy really would occur, and it is impractical to calculate, so we pick infinitely far away because that is what makes the math as simple as can be.
It comes from integrating G*M*m/r^2 relative to r, which produces a +C arbitrary constant of integration. Since most of the time, we are only interested in differences in potential energy, rather than absolute potential energy (wich doesn't really exist), we keep it simple and assign C=0 when defining potential energy as it derives from N's law of gravitation.
@@carultch Nice Logical Answer but ..
Both Situation ( Whether object are inside of the planet's finite Gravity field or Field with Infinite range from highest to lowest strength . ) Seems Violate the Conservation .
Take Hypothetical example for Gravity as finite strength -
Suppose Sun has Gravitational Field of 10AU unit and its very sharp boundary of it's G - field and there is an object with 1 kg mass , just meter away from sun's G- Field and it just need very little energy to push that object in side that sun's Gravity region ( 10AD assumed ) but because of Sun's gravitational acceleration that little triggering energy will end up millions of times stronger when that 1KG object will hit the sun .
It seem Funny thought experiment but was just for saying - it doesn't matter if Gravitational field slowly fade out over distance or never ends towards infinity .
@@omsingharjit There are planets that are farther than 10 AU from the sun in the solar system, so your premise is already false. 10 AU is about the orbital position of Saturn.
You have to go far beyond the edge of our galaxy to find a hard-stop to the sun's gravitational field. It does exist eventually, just not at the scale of 10 AU. Eventually, the expansion of the universe gets in the way, and causes space to be created between us and distant galaxies. Where enough space is being created, the gravitational waves cannot propagate fast enough to get to the distant galaxies, and such distant galaxies are no longer in gravitational contact with our own sun and our own galaxy.
It is lucky that we are alive at a time when we can discover this, because other than the select few local galaxies like Andromeda and the Magellanic Clouds, most galaxies are far enough away from us that they are out of gravitational contact, and will perpetually move away. Eventually, they will move farther away than the observable universe, and no receding galaxies will be visible to us as evidence of the universal law of gravitation.
How we account for the energy of universe expansion is still a mystery, and is a concept that cosmologists call dark energy.
Hey prof could you pls release the vid of conservation of linear momentum asap
The goat
1000000 thanks for you explanation
bravo Dave. goo joob
How does energy conservation apply to a fulcrum?
www.physicsforums.com/threads/law-of-the-lever-conservation-of-energy-or-angular-momentum.833317/
Would this be the primary reason a magnetic motor can never provide more energy then is put into it?
I always talk about conservation of energy when I'm pissed off about politics too. Lol... Jk, kinda.
If it wasn't for school, I would not be here.
wgt a pendelum does not hit a person's face without flinching. when released at the same point at a given distance?
why?
@@hajimajinja4941 Because energy is conserved. There is no energy added to the swinging pendulum, when it fell from the person's face, to the opposite end of the room, and back to the person's face. The pendulum can go no higher, than the point at which it started.
Time would need to run backwards, for the air drag and friction to do positive work on the swinging pendulum, which would speed it up. As crazy as this sounds, it is perfectly OK with the first law of thermodynamics. It is the second law of thermodynamics, that makes forces like friction and air drag slow objects down instead of speed them up.
Legend.
Все понятно спасибо
I bet everyone who watched this video tried hitting his hand on the table💀💀 convince me u didnt😅
Hi, is there any formula for pendulum lever length periods for the free fall pendulum from 12 o`clock when potential energy on 12 clocks is zero?
Free fall pendulum is an oxymoron. The bob is still supported by the constraint force in the rod and pivot.
There is no closed form equation for high amplitude pendulums, where you can't approximate sin(theta) as theta. You can solve it with an infinite series if you need to model it, but you cannot get an exact equation in closed form for the period. It also no longer undergoes simple harmonic motion, when you can no longer take credit for the approximation.
Hyperphysics has a large amplitude pendulum period calculator, that enables you to see this effect.
if the object is falling, that means g is negative and h is negative . Right? so that's why you got positive answer (the negative cancelled themselves) .. instead of negative ????????
I always get confused with g. sometimes it be positive and sometimes it be negative. Please help!
The reason its sign changes, is that you assign a sign convention and coordinate system as you see fit to solve the problem. In any event, the gravitational field g acts downward, and your assignment of signs needs to account for this, as the downward direction fits your coordinate system. Sometimes it is more convenient to assign positive to the downward direction, other times it is more convenient to assign positive to the upward direction.
I am stuck on a problem, this is because there is no velocity given at ALL. How could you find the velocity without having the initial information (No time, or acceleration of the object falling was given)? I did try using the conservation of energy formula yet the answer came out to be no solution!
I guess you have to assume that the initial velocity is 0
I think there is a problem with sig figs, should be 41 m/s instead
Never question the examiner
Ty
I have a question, how did you solve for its final velocity?
square root of 1,640 = 40.5
Using v² = u² + 2gh formula you can solve= it.... Here I am explaining....
v = ? , u = 40.5 m/s² , g = 9.8m/s²
h = 20 m
So, v² = (40.5)² + 2 × 9.8 × 20
= 2032.25
So, v = 45.08 m/s²
So the final velocity is 45.08 m/s²
you are very cool
you to