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Vladimir79 [104]
2 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]2 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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Answer:

final velocity = 0

because the train stoped

so,

  • acceleration = (v - u) ÷ t
  • acceleration = (0 - 22) ÷ 135
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5 0
3 years ago
A saddle weighs 250 newtons. The mass of the saddle is __________ kilograms. Use g = 9.8 N/kg for gravity.
Irina18 [472]

Answer:

saddle weighs 250 newtons. The mass of the saddle is ____250/9.8 kg______ kilograms. Use g = 9.8 N/kg for gravity.

7 0
2 years ago
Which object has the least momentum? object a: m = 1 kg, v = 100 m/s object b: m = 10 kg, v = 12 m/s object c: m = 0.5 kg, v = 1
nirvana33 [79]
A = 1*100 = 100 Ns
b = 10 * 12 = 120 Ns
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8 0
3 years ago
explain how a maglev train provides evidence that magnetic fields can apply forces between objects that do not touch.
Furkat [3]

Answer:

Powerful electromagnets are fitted on top of guideways either ttract or repel the magents fitted on the bottom of train. The froce of attraction/repulsion rasies the train in hovering position.

Explanation:

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4 0
2 years ago
A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co
Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = -F_Dd

Where:

d = Distance traveled = 450 m

Therefore,

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd and

v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh  -F_Dd}{ \frac{1}{2}m}  = v_1^2 + 2gh -\frac{   2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0
3 years ago
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