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The kinetic energy of a rolling body depends on:
In rolling motion, the total kinetic energy is the sum of translational 1/2𝑀𝑣² and rotational 1/2𝐼𝜔² kinetic energy. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particles and Rotational Motion. Give answer according to your understanding. For more pleasRead more
In rolling motion, the total kinetic energy is the sum of translational 1/2𝑀𝑣² and rotational 1/2𝐼𝜔² kinetic energy. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particles and Rotational Motion. Give answer according to your understanding.
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If a body is rotating with uniform angular velocity, which of the following quantities is zero?
For uniform angular velocity, there is no change in 𝜔 over time, so angular acceleration 𝛼 and torque 𝜏 are both zero. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational Motion Give answer according to your understanding. For more pleasRead more
For uniform angular velocity, there is no change in 𝜔 over time, so angular acceleration 𝛼 and torque 𝜏 are both zero. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational Motion Give answer according to your understanding.
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What is the center of mass of a uniform semicircular ring of radius 𝑅 ?
The center of mass of a uniform semicircular ring lies at a distance 2𝑅/𝜋 along the axis of symmetry, derived using integration over the arc. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your underRead more
The center of mass of a uniform semicircular ring lies at a distance 2𝑅/𝜋 along the axis of symmetry, derived using integration over the arc.
This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is
In the case of a frictionless inclined table, the work done by the table surface on the ball can be analyzed through the forces acting on the ball. Since there is no friction, the only force acting parallel to the surface is gravity. Gravity does not do work against the normal force of the table. ThRead more
In the case of a frictionless inclined table, the work done by the table surface on the ball can be analyzed through the forces acting on the ball.
Since there is no friction, the only force acting parallel to the surface is gravity. Gravity does not do work against the normal force of the table. The normal force acts perpendicular to the displacement of the ball.
Work done (W) is given by the formula:
W = F • d • cos(θ)
Where:
– F is the force
– d is the displacement
– θ is the angle between the force and displacement
This implies that the displacement and the force exerted are perpendicular to each other, that is, θ = 90 degrees. This gives cos(90°) = 0. Thus the work done by the table surface on the ball is:
W = F • d • 0 = 0
Final Answer:
The work done by the table surface on the ball is zero.
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A body of mass 10 kg initially at rest acquires velocity of 10 ms⁻¹. What is the work done?
To calculate the work done on the body, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. Step 1: Calculate the initial kinetic energy (K.E.₁) Since the body is initially at rest, its initial kinetic energy is: K.E.₁ = (1Read more
To calculate the work done on the body, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
Step 1: Calculate the initial kinetic energy (K.E.₁)
Since the body is initially at rest, its initial kinetic energy is:
K.E.₁ = (1/2) m v₁²
K.E.₁ = (1/2) × 10 kg × (0 m/s)²
K.E.₁ = 0 J
Step 2: Final kinetic energy K.E.₂
When the body has achieved a velocity of 10 m/s, the final kinetic energy is:
K.E.₂ = (1/2) m v₂²
K.E.₂ = (1/2) × 10 kg × (10 m/s)²
K.E.₂ = (1/2) × 10 × 100
K.E.₂ = 500 J
Step 3: Work done (W)
The work done equals the change in kinetic energy
W = K.E.₂ – K.E.₁
W = 500 J – 0 J
W = 500 J
Final Answer:
Work done is 500 J.
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