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3 years ago

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A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s

, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?

1 answer:

finlep [7]3 years ago

4 0

Answer:

$a=24.025\ m/s^2$

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

$a=15.5^2\times 0.1\ m/s^2$

$a=24.025\ m/s^2$

Therefore the acceleration of the clay will be $a=24.025\ m/s^2$ .

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A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

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Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a) $Ke=\frac{1}{2}mv_{max}^2$

$6.74=\frac{1}{2}\times 0.267\times v_{max}^2$

$v_{max}=7.1m/sec$

(b) Now $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec$

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$k=339.9N/m$

(c) We know that energy is given by

$E=\frac{1}{2}KA^2$

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$A=19.91cm$

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2 years ago

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Norma-Jean [14]

Answer:

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$T_{max} = T_{allow}( \frac{Ip}{r})$

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Answer:

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