Our values can be defined like this,
![m = 65kg](https://tex.z-dn.net/?f=m%20%3D%2065kg)
![v = 3.5m / s](https://tex.z-dn.net/?f=v%20%3D%203.5m%20%2F%20s)
![d = 0.55m](https://tex.z-dn.net/?f=d%20%3D%200.55m)
The problem can be solved for part A, through the Work Theorem that says the following,
![W = \Delta KE](https://tex.z-dn.net/?f=W%20%3D%20%5CDelta%20KE)
Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by
![\frac {1} {2} mv ^ 2](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%20%7B2%7D%20mv%20%5E%202)
So,
![Fd = \frac {1} {2} m ^ 2](https://tex.z-dn.net/?f=Fd%20%3D%20%5Cfrac%20%7B1%7D%20%7B2%7D%20m%20%5E%202)
Clearing F,
![F = \frac {mv ^ 2} {2d}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%20%7Bmv%20%5E%202%7D%20%7B2d%7D)
Replacing the values
![F = \frac {(65) (3.5)} {2 * 0.55}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%20%7B%2865%29%20%283.5%29%7D%20%7B2%20%2A%200.55%7D)
![F = 723.9N](https://tex.z-dn.net/?f=F%20%3D%20723.9N)
B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.
Resistance is the name of the ratio of the voltage applied to a circuit and the current in a circuit. Goes under <span>Ohm's Law</span>
Answer: Is there supposed to be something elses here because this question is unclear.
Explanation:
Answer:
V=IR
V=0.8×2.5
V=2V
where V=potential difference
I=Current
R=Resistance
.........