Science Net Forces. Could somebody help me?

Physics

1 answer:

tensa zangetsu [6.8K]3 years ago

7 0

2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.

3) Balanced. Both boys are pulling with the same force. Neither is winning.

4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.

5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.

6) Balanced. You and the dog are pulling with the same force.

You might be interested in

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

Veseljchak [2.6K]

a. The magnitude of the change of the momentum of the 5-kg ball is equal to the magnitude of the change of momentum of the 10-kg ball.

c. The magnitude of the change of velocity the 5-kg ball experiences is greater than that of the 10-kg ball.

Explanation:

For an inelastic collision:

The total momentum of the system is conserved
The total kinetic energy of the system is not conserved

Using these facts, let's now analyze each statement given.

a. The magnitude of the change of the momentum of the 5-kg ball is equal to the magnitude of the change of momentum of the 10-kg ball. --> TRUE. Since the total momentum is conserved, we can write:

$p_1 = p_1'+p_2'$

where

$p_1$ is the initial momentum of the 5-kg ball

$p_1'$ is the final momentum of the 5-kg ball

$p_2'$ is the final momentum of the 10-kg ball

The equation can be rewritten as

$p_1-p_1'=p_2'$

which is equivalent to

$-\Delta p_1 = \Delta p_2$

which means that the magnitude of the change of momentum of the two balls is the same.

b. Both balls lose all their momentum since the collision is inelastic. --> FALSE, the 10-kg ball gains momentum, so it does not lose it.

c. The magnitude of the change of velocity the 5-kg ball experiences is greater than that of the 10-kg ball. --> TRUE. We already said that the magnitude of the change in momentum of the two balls is the same. However, it can be written as

$\Delta p = m\Delta v$

where m is the mass of the ball and $\Delta v$ its change in velocity. Therefore, the 5-kg ball (which has smaller mass) will have a larger $\Delta v$ , so a larger change in velocity.