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Alla [95]
3 years ago
10

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Physics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2

the angular speed is given by:

\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s

to calculate how many radians have the wheel turned we need the apply the following formula:

\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad

the distance is given by:

d=\theta*r

d=381rad*(35.5*10^{-2}m)\\d=135m

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1. You released a pendulum of mass 1kg from a height of 0.05m
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a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

<h3>Kinetic energy of the pendulum when it reaches bottom</h3>

K.E = 100%P.E - 18%P.E

where;

  • P.E is potential; energy

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

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<h3>Speed of the pendulum</h3>

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

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v = √0.804

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<h3>Final potential energy </h3>

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

<h3>Height reached by the pendulum</h3>

P.E = mgh

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h = (0.374)/(1 x 9.8)

h = 0.038 m

<h3>when the pendulum stops</h3>

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Learn more about pendulum here: brainly.com/question/26449711
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