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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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Explanation:

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2 years ago

A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

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Answer:

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Explanation:

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Using mirror's formula :

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3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

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Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

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Where:

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Where:

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$U_{2} = m\cdot g \cdot \Delta y$

Where:

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The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

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$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

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3 years ago