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The correct answer is d) Spore formation. Vegetative propagation is a type of asexual reproduction in plants where new plants are produced from the vegetative parts of the plant such as roots, stems, or leaves. The key characteristic of vegetative propagation is that it does not involve seeds or spores. a) Cutting: This is a common method of vegetative propagation. A part of the plant, usually a stem or leaf, is cut and planted. This cut part can grow into a new plant. b) Layering: In this method, a branch of the plant is bent to the ground and covered with soil. After some time, roots develop from the covered part, and a new plant grows. c) Budding: This is another method of vegetative propagation where a bud from one plant is inserted into another plant. This technique is often used in horticulture to propagate plants like roses. d) Spore formation: This is not a method of vegetative propagation. Spore formation is a method of asexual reproduction seen in fungi, algae, and some plants like ferns. Spores are microscopic, reproductive units that can develop into a new organism. Summary: The three methods-cutting, layering, and budding-are all techniques of vegetative propagation. Spore formation, however, is a different process used by certain organisms for reproduction, not a method of vegetative propagation in plants.

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The correct answer is (a) 10 Sodium (Na) has an atomic number of 11, meaning it has 11 protons and, in its neutral state, 11 electrons. When a sodium atom loses one electron, it becomes a sodium ion (Na⁺). The loss of one electron reduces the total number of electrons from 11 to 10. Therefore, the Na⁺ ion has 10 electrons.

I am a teacher so I look at the book for questions like this

@@studyRoom31 great job, but I asking about the video editing tools.

@@MuhammadAli-dy6kj Yeah, it requires an app and that app is called (Kine Master)

@@studyRoom31 okay thanks

@@MuhammadAli-dy6kj triple tanx

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The percentage error in a measurement quantifies how much the measured value deviates from the true value, expressed as a percentage of the true value. The formula to calculate the percentage error is: Percentage Error=(∣True Value−Measured Value∣True Value)×100\text{Percentage Error} = \left( \frac{|\text{True Value} - \text{Measured Value}|}{\text{True Value}} ight) \times 100Percentage Error=(True Value∣True Value−Measured Value∣)×100 Given: True Value = 2.60 m Measured Value = 2.58 m Let's plug these values into the formula: Percentage Error=(∣2.60−2.58∣2.60)×100\text{Percentage Error} = \left( \frac{|2.60 - 2.58|}{2.60} ight) \times 100Percentage Error=(2.60∣2.60−2.58∣)×100 First, find the absolute difference:

@@wordsinpaperjournaljournal 2.60−2.58∣=0.02|2.60 - 2.58| = 0.02∣2.60−2.58∣=0.02 Next, divide this difference by the true value: 0.022.60≈0.007692\frac{0.02}{2.60} \approx 0.0076922.600.02≈0.007692 Finally, multiply by 100 to get the percentage: 0.007692×100≈0.76920.007692 \times 100 \approx 0.76920.007692×100≈0.7692 So, the percentage error is approximately: Percentage Error≈0.77%\text{Percentage Error} \approx 0.77\%Percentage Error≈0.77%

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Determine the net displacement direction and distance: Moving 10 meters north is a positive displacement in the north direction. Moving 5 meters south is a negative displacement in the north direction. Calculate the net displacement: Net displacement = 10   meters north − 5   meters south = 5   meters north Net displacement=10meters north−5meters south=5meters north Magnitude of the displacement: The magnitude of the displacement is the absolute value of the net displacement. Magnitude of displacement = ∣ 5   meters ∣ = 5   meters Magnitude of displacement=∣5meters∣=5meters Thus, the magnitude of the displacement is 5 meters.

Correct answer option A

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Yeah sure.It is the correct answer option A

I am so sorry Brother, I don't understand what you are talking about, explain to me a little so that I can answer you.

@@studyRoom31 sir g my question is that in your question is confusion b/c you talking about the speed and acceleration not centrepetal acceleration plz don`t mind and check your answer

@@syedsaliman2195 Acceleration is defined as the rate of change of velocity with respect to time. Velocity includes both speed and direction. If an object is moving at a constant speed, its speed does not change. However, for the object to have zero acceleration, its direction must also remain constant. Thus, when an object moves at a constant speed in a straight line (constant velocity), there is no change in its velocity, and hence its acceleration is zero. In summary: Constant speed implies no change in the magnitude of velocity. For acceleration to be zero, the direction must also be constant. Therefore, constant speed in a straight line results in zero acceleration.

Sir, but there in the question it does not mention that the body, is traveling around the circumference of circle. Hence answer will be zero.​@@studyRoom31

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Yeah sure! To determine the acceleration of an object when a force is applied, you can use Newton's second law of motion, which is given by the formula: F=ma F=ma where: F F is the force applied, m m is the mass of the object, a a is the acceleration of the object. Given: Force F=10 N F=10N Mass m = 5 kg m=5kg We need to find the acceleration a a. Rearranging the formula to solve for a α: a=Fm α = m F Substituting the given values: a=10 N5 kg α = 5kg 10N a=2 m/s2 a=2m/s 2 So, the acceleration of the object is 2 m/s2 2m/s 2

Yes it's wrong

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I don't know

Yeah sure!To determine the acceleration of an object when a force is applied, you can use Newton's second law of motion, which is given by the formula: 𝐹=𝑚𝑎 F=ma where: 𝐹 F is the force applied, 𝑚 m is the mass of the object, 𝑎 a is the acceleration of the object. Given: Force 𝐹=10 N F=10N Mass 𝑚=5 kg m=5kg We need to find the acceleration 𝑎 a. Rearranging the formula to solve for 𝑎 a: 𝑎=𝐹𝑚 a= m F Substituting the given values: 𝑎=10 N5 kg a= 5kg 10N 𝑎=2 m/s2 a=2m/s 2 So, the acceleration of the object is 2 m/s2 2m/s 2 .

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Which answer is wrong tell me to see the book quickly?

Ans.of 12th qus. Wrong 2m/ss is correct

Ok