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uranmaximum [27]
3 years ago
9

A 16 lb weight stretches a spring 6 inches in equilibrium. It is attached to a damping mechanism with constant c. Find all value

s of c such that the free vibration of the weight has infinitely many oscillations.
Physics
1 answer:
arlik [135]3 years ago
6 0

Answer:

-8

Explanation:

Given data:

Weight W = 16 lb

length l = 6/12 = 0.5 ft

hence, spring constant  k = W/l = 16/0.5 = 32 lb/ft

The equation of motion of spring is

mx''+cx+kx=0\\0.5x''+cx'+32x=0

the auxiliary equation can be written as

0.5r^2+cr+32=0\\r^2+2cr+64=0

The discriminate of equation is

D =(2c)^2+-4(1)(64)\\=4(c^2-64)

To get the value of the damping constant,

D

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Explanation:

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6 0
2 years ago
One speaker generates sound waves with amplitude A.
raketka [301]

Answer:

iv) It is 9x bigger than before

Explanation:

As the amplitudes of the new speakers add directly with the original one, taking into account the phase that they have, the composed amplitude of the sound wave is as follows:

At = A + 4A -2A = 3 A

The intensity of the wave, assuming it propagates evenly in all directions, is constant at a given distance from the source, and can be expressed as follows:

I = P/A

where P= Power of the wave source, A= Area (for a point source, is equal to the surface area of a sphere of radius r, where is r is the distance to the source along a straight line)

For a sinusoidal wave, the power is proportional to the square of the amplitude, so the intensity is proportional to the square of the amplitude also.

If the amplitude changes increasing three times, the change in intensity will be proportional to the square of the change in amplitude, i.e., it will be 9 times bigger.

So, the statement iv) is the right one.

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3 years ago
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A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

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8 0
3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo
Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = 83-44.3=38.7^{\circ}

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, d^2 representing the distance between the planes, we see that:

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d = 1324.9453 m

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3 years ago
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