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Nataly_w [17]
3 years ago
7

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts

to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.25 m/s2, determine the coefficient of static friction between the trunk and the turntable.
Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

0.26

Explanation:

The radius of the table is 2.5 m.

Tangential velocity,

v=u+at\\u=0\\t=10s\\a=0.25m/s^2\\v=0+0.25\times 10 = 2.5 m/s

Net force on the trunk in the plane of the table:

F=m\sqrt{a^2+(\frac{v^2}{r})^2}\\F= m\sqrt{0.25^2+(\frac{2.5^2}{2.5})^2}=2.5m

The trunk slides because the frictional force equals the net force acting on the table:

\mu m g = 2.5 m\\\mu =\frac{2.5}{g} = 0.26

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Answer:

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Explanation:

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Only the specific heat capacity (per unit mass) of the solution is given. Both the mass of the solution and the temperature change will be required for determining the energy change. Start by finding the mass of the solution.

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The heat capacity of the coffee cup is given. Only the temperature change will be required for finding the amount of heat absorbed.

Q = C\cdot \Delta T = \rm 30.7\times 9.915 = 304.391\; J.

<h3>Heat that this reaction produces</h3>

Find the sum of the two parts of heat. Round to three significant figures as in the heat capacity of the coffee cup and the density of the solution.

\rm 1335.80+ 304.391 = 1.64\times 10^{3}\; J.

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f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}} = \frac{110 Hz}{\sqrt{\frac{594 N}{554 N}}} = 106.23 Hz

Hence, the beat frequency is:        

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Therefore, the beat frequency that is heard when the hammer strikes the two strings simultaneously is 3.77 Hz.

I hope it helps you!                              

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