Answer:
Explanation:
<h3>Given Data:</h3>
Mass = m = 68 kg
Velocity = v = 30 m/s
Time = 2 hours = 2 × 60 × 60 = 7200 s
<h3>Required:</h3>
Force = F = ?
<h3>Formula to be used:</h3>
<h3>Solution:</h3>
![\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%20%3D%20%5Cfrac%7B%2868%29%2830%29%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B2040%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%200.28%20N%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0
~Formula: Voltage= current• resistance
(V= Ir)
~Using this formula, plug in the numbers from the equation into the formula
~5=25i
~Now you have a one-step equation
~Divide by 25 on both sides and you should get your answer:
~I= 0.2 (which means current is 0.2)
