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larisa86 [58]
2 years ago
13

g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m

oves offers a damping force numerically equivalent to92times the instantaneous velocity.(a) Find the equation of motion if the mass is initially released from the equilibrium position with a downwardvelocity of 2 ft/s. (Use the convention that displacements measured below the equilibrium position are positive).(b) What is the velocity of the mass whent
Physics
1 answer:
marta [7]2 years ago
3 0

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo)  = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

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A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity
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The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

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3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
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Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

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A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

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3 years ago
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