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garri49 [273]
3 years ago
13

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

irection at 1.40 m/s. All cars have a mass of mass 1.6 104 kg. Determine the following.(a) speed of the three coupled cars after the collision (Give your answer to at least 2 decimal places.) m/s(b) kinetic energy lost in the collision J
Physics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

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Explanation :

It is given that,

Mass of the car, m = 1000 kg              

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\dfrac{F_A-\mu mg}{m}=a

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Answer:

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Explanation:

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.

If the muscle generates a force

F =  49.5 N and r = 2.65 cm , then the torque is equal to rF

we see that r = 2.65 cm = 0.0265 m

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torque = 0.0265 x 49.5

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car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
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Answer:

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Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

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Ff = √(F2ft +F2fc)= matp √(1+π²)

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1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

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