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oksano4ka [1.4K]
2 years ago
5

If you were trying to build a soundproof room, which of the following materials would you choose in order to absorb the most sou

nd, based on the coefficient of absorption for each material?
A. Heavy curtains
B. Wood
C. Concrete
D. Carpet
Physics
2 answers:
leonid [27]2 years ago
5 0

Explanation :  

Absorption coefficient of a material determines how much sound is absorbed by the material.

To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.

Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.  

Hence, option (A) and (D) are correct.

Margarita [4]2 years ago
5 0
Carpet and heavy curtains, and wood. concrete is more liable to echo. with the heavy carpet, its thick enough to weaken the sounds, as well as carpet. wood is thick and big enough to do the same aswell. hope this helps.
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Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

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W_1 = 57.96 J

Now similarly work done by frictional force

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Now total work done on sled

W_{net}= W_1 + W2

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B. there would be a global rise in temperatures

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T
atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.

<em>where</em>,

k is the Coulomb's constant, having value = \dfrac{1}{4\pi \epsilon_o}

\epsilon_o is the electrical permittivity of free space.

Also, r>>s, therefore, \rm r^2+s^2\approx r^2.

\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.

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\rm F=qE

Therefore, the electric force on the charge Q due to the dipole is given by

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According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

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Since the charge Q is placed in the plane that bisects the dipole, therefore, \theta = 90^\circ.

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2 years ago
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