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elena-14-01-66 [18.8K]
3 years ago
12

A wheel rolling on a horizontal surface with an angular speed of 2.5 rad/s gets on a ramp and rolls down the ramp with a constan

t angular acceleration of 2.0 rad/s2. If it takes 11.5 us to reach the bottom of the ramp, what is the final angular speed of the wheel at the bottom
Physics
1 answer:
STALIN [3.7K]3 years ago
7 0

Answer:

the final angular speed of the wheel at the bottom is 25.5 rad/s

Explanation:

The computation of the final angular speed of the wheel at the bottom is as follows:

As we know that

W_f = W_{in} + \alpha t\\\\= 2.5 + (2 \times 11.5)\\\\= 2.5 + 23\\\\= 25.5 rad/s

Hence, the final angular speed of the wheel at the bottom is 25.5 rad/s

We simply applied the above formula so that the final angular speed could come

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Read 2 more answers
A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

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3 years ago
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