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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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Answer:

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A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas
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3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

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f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

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