Answer:
d = 2.54 [m]
Explanation:
Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

where:
W = work [J]
Ek1 = kinetic energy at initial state [J]
Ek2 = kinetic energy at the final state = 0.
We must remember that kinetic energy can be calculated by means of the following expression.
![\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D-W%3D0%5C%5CW%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A4%2A%285%29%5E%7B2%7D%5C%5CW%3D%2050%20%5BJ%5D)
We know that work is defined as the product of force by distance.

where:
F = force [N]
d = distance [m]
But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.
![f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]](https://tex.z-dn.net/?f=f%3Dm%2Ag%2A0.5%5C%5Cf%20%3D%204%2A9.81%2A0.5%5C%5Cf%20%3D%2019.62%20%5BN%5D)
Now solving the equation for the work.
![d=W/F\\d = 50/19.62\\d = 2.54[m]](https://tex.z-dn.net/?f=d%3DW%2FF%5C%5Cd%20%3D%2050%2F19.62%5C%5Cd%20%3D%202.54%5Bm%5D)
Answer:
<em>The person needs to apply 25 N to balance the seesaw</em>
Explanation:
<u>Moment</u>
The moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis.
The moment M of a force F located at a distance x from the axis of rotation is calculated as follows:
M = F.x
The image shows a moment of M=100 N.m is needed to be applied to balance the seesaw. It can also be noted that the distance to the pivot is x=4 m
To calculate the force needed to balance the seesaw, we solve for F:


F = 25 N
The person needs to apply 25 N to balance the seesaw
Answer:
160 kg
12 m/s
Explanation:
= Mass of first car = 120 kg
= Mass of second car
= Initial Velocity of first car = 14 m/s
= Initial Velocity of second car = 0 m/s
= Final Velocity of first car = -2 m/s
= Final Velocity of second car
For perfectly elastic collision

Applying in the next equation


Mass of second car = 160 kg
Velocity of second car = 12 m/s
D.
The reading between 7N and 8N would have to be 7.5N. Answers A and B are much to small and answer C is way to big.