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ollegr [7]
3 years ago
8

A compressed spring is compressed between a 5kg and 3 kg cart, respectively. When the spring is released, what will be the same

for the carts?
Physics
1 answer:
stira [4]3 years ago
4 0

The momentum of the two carts will be the same

Explanation:

First of all, we notice that the system consisting of the two carts + the spring is an isolated system: this means that there are no external forces acting on it.

As a result, the total momentum of the system must be conserved.

Since the spring is at rest, the total momentum of the system after the two carts are released is:

p_f = p_1 + p_2

where

p_1 is the momentum of the 1st cart

p_2 is the momentum of the 2nd cart

The total momentum of the system at the beginning, when the spring is compressed, is zero:

p_i=0

Since the total momentum must be conserved, we have:

p_i = p_f\\0 = p_1 + p_2\\\rightarrow p_2 = -p_1

Which means that the two carts will have equal and opposite momenta.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
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Answer:

The angular velocity is 15.37 rad/s

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To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

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8 0
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A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
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Answer:

a) \Delta U_g=-5.3kJ

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U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

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the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

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the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

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