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MariettaO [177]
2 years ago
9

You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical

loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s.
Physics
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

Work done by friction along the motion is given as

W_f = -5857.8 J

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)

W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)

now plug in all data in it

W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)

W_f = 30607.2 - 36465

W_f = -5857.8 J

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Answer:

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Explanation:

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A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
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The Young modulus is given by:
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where
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A is the cross-sectional area of the wire
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We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
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Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

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Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

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Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

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Case (3)

q_x  =-50*(160)*10^3==>-8 kW

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Case (4)

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Case (5)

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note:

all graph are attached

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Answer:

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Explanation:

Given:

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