Answer:
A rectangular object may have different areas for different sides.
If that object is placed on a hand, the area of its side does not affects the Force. However the pressure it puts on the hand is affected by the area.
Explanation:
The Force that object applies on the hand is given as the product of its mass and gravitational acceleration 'g'. Hence area does not affect the Force and it is constant.
F = mg
Pressure is defined as the Force per unit area.
P = F/A
As the surface area decreases and Force remains constant, the pressure on the hand increases and vice versa.
Answer:
,,,,,,
Explanation:
m jkg yrfyuih iopoj kpoohjkbhj huiph i
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
Answer:
1.8*10^8m/s
Explanation:
Using
L= lo√1-v²/c²
So making v subject we have
V= c√1-4.8²/6²
V= 0.6*c
V= 0.6*3E8m/s
V= 1.8*10^8m/s
Your answer is ''Uniform''.
Hope this helps :)
