Answer:
W = 28226.88 N
Explanation:
Given,
Mass of the satellite, m = 5832 Kg
Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m
The time period of the orbit, T = 1.9 h
= 6840 s
The radius of the planet, R = 4.38 x 10⁶ m
The time period of the satellite is given by the formula
second
Squaring the terms and solving it for 'g'
g = 4 π²
m/s²
Substituting the values in the above equation
g = 4 π²
g = 4.84 m/s²
Therefore, the weight
w = m x g newton
= 5832 Kg x 4.84 m/s²
= 28226.88 N
Hence, the weight of the satellite at the surface, W = 28226.88 N
Answer:
Pressure, P = 32666.66 Pa
Explanation:
It is given that,
Surface area of foot of Bimaba is 150 cm² or 0.015 m².
Her weight is 50 kg
We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :

So, the pressure is 32666.66 Pa.
Answer:
Explanation:bjbhfcvvjkkknbnnnm
Answer:
(A) 60 J
Explanation:
At state 1
KE₁=100 J
At state 2
KE₂ = 0
U₂=80 J
Given that surface is rough so friction force will act in opposite to the direction of motion
Lets take work done by friction = Wfr
From work power energy
Work done by all forces = Change in kinetic energy
Wfr + U₂=ΔKE
Wfr+80 = 100
Wfr= 20 J
Now when book slides from top position then
Wfr+ U = KEf - KEi
-20 + 80 = KEf-0
KEf= 60 J
(A) 60 J