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NISA [10]
3 years ago
6

The cable of a hoist has a cross-section of 80 mm 2 . The hoist is used to lift a crate weighing 500 kg. What is the stress in t

he cable? The free length of the cable is 3 m. How much will it extend if it is made of steel (modulus 200 GPa)? How much if it is made of polypropylene, PP (modulus 1.2 GPa)?
Physics
1 answer:
motikmotik3 years ago
4 0

Answer:

0.0009196875 m

0.15328125 m

Explanation:

Y = Young's modulus

Stress is given by

\sigma=\dfrac{mg}{A}\\\Rightarrow \sigma=\dfrac{500\times 9.81}{80\times 10^{-6}}\\\Rightarrow \sigma=61312500\ Pa

Change in length is given by

\Delta L=\dfrac{L\sigma}{Y}\\\Rightarrow \Delta L=\dfrac{3\times 61312500}{200\times 10^9}\\\Rightarrow \Delta L=0.0009196875\ m

The change in length of steel is 0.0009196875 m

\Delta L=\dfrac{L\sigma}{Y}\\\Rightarrow \Delta L=\dfrac{3\times 61312500}{1.2\times 10^9}\\\Rightarrow \Delta L=0.15328125\ m

The change in length of polypropylene is 0.15328125 m

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hodyreva [135]

Answer:

a) 1.517\times10^{-11} s

b) 3.41 mm

Explanation:

a)

We take the speed of light, c = 3.0\times10^8 m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

n=\dfrac{c}{s}

s=\dfrac{c}{n}

t=\dfrac{d}{c/n}

t=\dfrac{dn}{c}

t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}

t=1.517\times10^{-11}

b)

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

d=s\times t

d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}

d=\dfrac{3\times10^{-3}\times 1.517}{1.333}

d = 3.41 mm

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3 years ago
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While driving a 2150 kg car, carl steps on the gas pedal, accelerating at 4.0 m/s2. If the coefficient of friction between his c
Mrac [35]

Answer:

d. 3332.5 [N]

Explanation:

To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.

Here we have two forces, the force that pushes the car to move forward and the friction force.

The friction force is equal to the product of the normal force by the coefficient of friction.

f = N * μ

f = (m*g) * μ

where:

N = weight of the car = 2150*9.81 = 21091.5 [N]

μ = 0.25

f = (21091.5) * 0.25

f = 5273 [N]

Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:

F + f = m*a

F + 5273 = 2150*4

F = 8600 - 5273

F = 3327 [N]

Therefore the answer is d.

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3 years ago
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