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A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10-4J by the end of the second revolution after the beginning of the motion ?
- a)0.2 m/s2
- b)0.1 m/s2
- c)0.15 m/s2
- d)0.18 m/s2
Correct answer is option 'B'. Can you explain this answer?
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To find the magnitude of the acceleration, we can use the equation for kinetic energy:
KE = 1/2mv^2
where KE is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.
Since the particle is moving along a circle, we can relate the velocity to the radius and the angular velocity. The angular velocity, ω, is related to the linear velocity, v, by the equation:
v = ωr
where r is the radius of the circle.
We are given that the particle completes two revolutions, so we can find the angular displacement, θ, using the equation:
θ = 2πn
where n is the number of revolutions.
Using the equation for angular displacement, we can find the angular velocity:
ω = θ/t
where t is the time taken to complete the two revolutions.
We are also given that the kinetic energy at the end of the second revolution is 8 x 10^-4 J. Therefore, we can write the equation:
8 x 10^-4 J = 1/2mv^2
Substituting v = ωr and θ = 2πn, we can rewrite the equation as:
8 x 10^-4 J = 1/2m(2πn)^2r^2
Simplifying the equation, we get:
8 x 10^-4 J = 2π^2n^2mr^2
Now, we need to find the acceleration, a. We know that acceleration is the rate of change of velocity, which can be written as:
a = Δv/Δt
Since the particle has a constant tangential acceleration, we can write:
a = v/t
Substituting v = ωr and θ = 2πn, we can rewrite the equation as:
a = ωr/t
Now, we need to find the time taken, t. We know that the time taken to complete two revolutions is equal to the period of the motion, T, multiplied by 2. The period, T, is the time taken to complete one revolution and is given by:
T = 1/f
where f is the frequency of the motion. The frequency, f, is the number of revolutions per second and is given by:
f = 1/T
Substituting the values for n and T, we can find the time taken, t:
t = 2/f = 2T = 2(1/f) = 2(1/(1/T)) = 2T = 2(2πn)
Substituting this value of t into the equation for acceleration, we get:
a = ωr/t = ωr/(2(2πn)) = ωr/(4πn)
Now, we can substitute the value of ωr from the equation for kinetic energy:
a = (2πn)(r/t)/(4πn) = r/2t
Now, we can substitute the values for r and t:
a = (6.4 cm)/(2(2πn)) = (6.4 cm)/(4πn) = 0.1 cm/s^2
Converting cm/s^2 to m/s^2, we get:
a = 0.1 m/s^2
Therefore, the magnitude
KE = 1/2mv^2
where KE is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.
Since the particle is moving along a circle, we can relate the velocity to the radius and the angular velocity. The angular velocity, ω, is related to the linear velocity, v, by the equation:
v = ωr
where r is the radius of the circle.
We are given that the particle completes two revolutions, so we can find the angular displacement, θ, using the equation:
θ = 2πn
where n is the number of revolutions.
Using the equation for angular displacement, we can find the angular velocity:
ω = θ/t
where t is the time taken to complete the two revolutions.
We are also given that the kinetic energy at the end of the second revolution is 8 x 10^-4 J. Therefore, we can write the equation:
8 x 10^-4 J = 1/2mv^2
Substituting v = ωr and θ = 2πn, we can rewrite the equation as:
8 x 10^-4 J = 1/2m(2πn)^2r^2
Simplifying the equation, we get:
8 x 10^-4 J = 2π^2n^2mr^2
Now, we need to find the acceleration, a. We know that acceleration is the rate of change of velocity, which can be written as:
a = Δv/Δt
Since the particle has a constant tangential acceleration, we can write:
a = v/t
Substituting v = ωr and θ = 2πn, we can rewrite the equation as:
a = ωr/t
Now, we need to find the time taken, t. We know that the time taken to complete two revolutions is equal to the period of the motion, T, multiplied by 2. The period, T, is the time taken to complete one revolution and is given by:
T = 1/f
where f is the frequency of the motion. The frequency, f, is the number of revolutions per second and is given by:
f = 1/T
Substituting the values for n and T, we can find the time taken, t:
t = 2/f = 2T = 2(1/f) = 2(1/(1/T)) = 2T = 2(2πn)
Substituting this value of t into the equation for acceleration, we get:
a = ωr/t = ωr/(2(2πn)) = ωr/(4πn)
Now, we can substitute the value of ωr from the equation for kinetic energy:
a = (2πn)(r/t)/(4πn) = r/2t
Now, we can substitute the values for r and t:
a = (6.4 cm)/(2(2πn)) = (6.4 cm)/(4πn) = 0.1 cm/s^2
Converting cm/s^2 to m/s^2, we get:
a = 0.1 m/s^2
Therefore, the magnitude
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A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer?
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A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer?.
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to8 x 10-4J by the end of the second revolution after the beginning of the motion ?a)0.2 m/s2b)0.1m/s2c)0.15 m/s2d)0.18 m/s2Correct answer is option 'B'. Can you explain this answer?.
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