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densk [106]
2 years ago
5

The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg

. For a very short time interval around 2. 0 s, an external force is exerted on an object in the system. What is the resulting change in momentum of the system?.
Physics
1 answer:
kumpel [21]2 years ago
6 0

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m

The change in the momentum is;

\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \  N.s

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
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Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

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\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

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