<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>
da answer is liquiddddddddd
A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
<span>The given values are
F1 = 200 N</span>
F2 =?
Total = 400 N
Solution:
F1 + F2 = T
200 N + F2 = 400N
F2 = 400 - 200
F2 = 200
N
Answer:

Explanation:
m = Masa del coche
g = Aceleración debida a la gravedad = 
h = Altura = 
v = Velocidad del automóvil en la parte inferior de la pista
Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

La velocidad máxima que puede alcanzar el coche es
.