Force of earth gravity on 1 kg mass is 9.8 N.
Since the mass is at earth surface, therefore its distance from the centre of earth is equal to the radius of the earth i.e. 6400 km.
The force of earth gravity is calculated as
F=GMm/r^2 =(6.67*〖10〗^(-11)*6*〖10〗^24*1)/〖(64*〖10〗^6)〗^2 =9.8 N
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Answer:using Newton third law
Let initial velocity of block be u1=3m/s
Mass of moving block m1 =1kg
Final velocity of block =V
Mass of stationary block m2= 4kg
Since they stick together, their final velocity will be the same.
m1u1 + m2u2=(m1+m2)v
(1*3)+(0*4)=(1+4)v
3=5v
Divide both sides by 5
V=0.6
Final velocity is 0.6m/s
Explanation:
Answer:
Explanation:
a. Landing height is
H=1.3m
Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft
u=1.3m/s
Velocity of lander at impact, i.e final velocity is needed
v=?
The acceleration due to gravity is 0.4 times that of the one on earth,
Then, g on earth is approximately 9.81m/s²
Then, g on Mars is
g=0.4×9.81=3.924m/s²
Then using equation of motion for a free fall body
v²=u²+2gH
v²=1.3²+2×3.924×1.3
v²=1.69+10.2024
v²=11.8924
v=√11.8924
v=3.45m/s
The impact velocity of the spacecraft is 3.45m/s
b. For a lunar module, the safe velocity landing is 3m/s
v=3m/s.
Given that the initial velocity is 1.2m/s²
We already know acceleration due to gravity on Mars is g=3.924m/s²
The we need to know the maximum height to have a safe velocity of 3m/s
Then using equation of motion
v²=u²+2gH
3²=1.2²+2×3.924H
9=1.44+7.848H
9-1.44=7.848H
7.56=7.848H
H=7.56/7.848
H=0.963m
The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m
Answer:
Tension in string equals 14.715 Newtons
Explanation:
The situation is represented in the figure attached below:
For equilibrium of the clothes along y- axis we have

Applying values we get
