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WARRIOR [948]
3 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]3 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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The output force is given by;

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The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2

Therefore, the cross-sectional area of the larger piston is 392 cm²

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ryzh [129]

Answer:

B. 2.3 x 10⁶

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