<u><em>Correct question:</em></u><u><em>a </em></u><u><em>200 g </em></u><u><em>ball is dropped from a height of 2m, bounces on a hard floor and rebounds to a height of 1.5m. What maximum force does the floor exert on the ball?</em></u>
<u><em>The diagram of the question is in the attachment.</em></u>
Answer:
Explanation:
V=√2gh
g-10m/s²
let u be initial velocity ad v be final velocity,
u=√2*10*2
u=6.324m/s
v=√2*10*1.5
v=5.477m/s
from the diagram
t=5ms=0.005s
F=-33.87N (the negative shows direction)
From the diagram Fmax=2F
Fmax= 2*33.87
=67.74N
Answer:
d = vi * t + ½ * a * t^2, a = -0.5
d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m
d6 = 20.8 * 6 – ½ * 0.5 * 6^2 = 106.8m
d6 – d5 = 15.3 m
Explanation:
Solution is in attachment ~
I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~
Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
Answer:
The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object: T = mg.