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WARRIOR [948]
2 years ago
5

The spin-drier of a washing machine revolving at 900.RPM slows down uniformly to 300.RPM while making 60. revolutions. Find the

angular acceleration and the time required to turn through these 60. revolutions.
Physics
1 answer:
jeyben [28]2 years ago
3 0
<h2>Answer:</h2>

<u>Acceleration is </u><u>-10.57 rad/s²  </u>

<u>Time is </u><u>6 seconds</u>

<h2>Explanation:</h2><h3>a) </h3>

u=900rpm= 94.248 rad/s  

v =300rpm= 31.416 rad/s  

s=60 revolutions= 377 rad  

v²= u²+ 2as  

31.416² = 94.248²+ 2 * a * 377  

a = v²-u² / 2s

a= -10.57 rad/s²  

<h3>b) </h3>

Using 1st equation of motion

v-u/a = t

Putting the values

t = (31.4 - 94.2)/-10.57

t = 6 seconds

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In your frame, a person is moving in the positive x-direction at a speed of 40,000 km / s. Suppose that a streetlamp located at
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Answer:

proper time taken by the person is 9.911 × 10⁻⁵ s

Explanation:

speed of the person  in x- direction = 40,000 km/s

                                                           =  40,000 × 10³ m/s

                                                           =  4 × 10⁷ m/s

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Lorentz factor = \gamma = \dfrac{1}{\sqrt{1-\dfrac{r^2}{c^2}}}

                                         = \dfrac{1}{\sqrt{1-\dfrac{(4 \times 10^7)^2}{(3 \times 10^8)2}}}

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time taken in your frame of reference,t =\dfrac{D}{v}

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hence, proper time taken by the person is 9.911 × 10⁻⁵ s

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