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Vladimir79 [104]
3 years ago
6

A person hangs from a nylon rope (Young's modulus of 5 x 109 N/m2) as seen in the picture below. The rope stretches by 2 % and h

as a diameter of 0.03 m. What mass does the person have to stretch the rope by this amount? (The answer may not be realistic!)
Physics
1 answer:
disa [49]3 years ago
8 0

Answer:

959183.7 kg  

Explanation:

from the question we have :

young modulus = 5 x 10^{9} N/m^{2}

strain = 2% = 2÷100 = 0.02

diameter = 0.03 m

radius = 0.015 m

acceleration due to gravity (g) = 9.8 m/s^{2}

we can get the mass from the formula below

young modulus = stress ÷ strain

where

stress = \frac[force}{area} = \frac {mass x g}{area}

area = 2πr = 2π x 0.015 = 0.094

therefore    

young modulus = \frac{\frac {mass x g}{area}}{strain}

 5 x 10^{9}  =  \frac{\frac {mass x 9.8}{0.094}}{0.02}

mass =  \frac{5 x 10^{9} x 0.02 x 0.094}{9.8}

mass = 959183.7 kg  

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A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m
charle [14.2K]

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) x(t) = (0.15m)\cos(2\pi (2.0Hz)t)

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

K_{eq} + U_{eq} = K_A + U_A\\\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\v = \sqrt{\frac{k}{m}}A

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Therefore,

v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s

Part B

The conservation of energy will be used again.

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}

where k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

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Part D

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x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)

where \phi is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

x(t) = (0.15m)\cos(2\pi (2.0Hz)t)

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4 years ago
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Answer:  

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6 0
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Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

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Explanation:

Part A

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The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

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t = The time taken to travel the distance, h = 5.2 s

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The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

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∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

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