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Olin [163]
3 years ago
10

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the

cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s. Through how many revolutions did the wheel rotate during the 5.0 seconds of braking?
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

\theta= \frac{\omega_f-\omega_i}{2}\times t

putting values

\theta= \frac{0-2}{2}\times5 = 5 rad

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I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME
yuradex [85]

Answer:

The total distance is  130.2 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the cart starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

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With this velocity we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

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The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} -(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

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v_{f} ^{2} =v_{o} ^{2} -2*a*x

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x = distance traveled [m]

21.6² = 24² - (2*1.6*x)

x = 109.44/(3.2)

x = 34.2 [m]

Note: The negative sign in the equations is because the car is desaccelerating, it means its velocity is decreasing.

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