Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease
As we know that sum of kinetic energy and potential energy will always remain conserved
So here we will have
so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.
So the pair of potential energy and kinetic energy will satisfy the above condition
The average force applied to the ball= 106.7 N
Explanation:
Force is given by
f= ΔP/t
ΔP= change in momentum= m Vf- m Vi
m= mass =0.2 kg
Vf= final velocity= 12 m/s
Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)
t= time= 60 ms= 0.06 s
now ΔP= 0.2 [ 12-(-20)]
ΔP=0.2 (32)=6.4 kg m/s
now force F= ΔP/t
F= 6.4/0.06
F=106.7 N
Answer:
g' = 13.5 m/s²
Explanation:
The acceleration due to gravity on surface of earth is given by the formula:
g = GMe/Re² --------------- euation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
Me = Mass of Earth
Re = Radius of Earth
Now, the the acceleration due to gravity on the surface of Kepler-62e is:
g' = GM'/R'² --------------- euation 1
where,
g' = acceleration due to gravity on surface of Kepler-62e
G = Universal Gravitational Constant
M' = Mass of Kepler-62e = 3.57 Me
R' = Radius of Kepler-62e = 1.61 Re
Therefore,
g' = G(3.57 Me)/(1.61 Re)²
g' = 1.38 GMe/Re²
using equation 1:
g' = 1.38 g
where,
g = 9.8 m/s²
Therefore,
g' = 1.38(9.8 m/s²)
<u>g' = 13.5 m/s²</u>
