First:
d = 100 m
t = 200 sec
v = 100/200 = 0.5 m/s
Displacement is zero since he returned to his start point.
t2 = d/v2 = 100/2 = 50 sec
total time = 50 + 200 + 500 = 750 sec
Answer:
a) A = 0.603 m
, b) a = 165.8 m / s²
, c) F = 331.7 N
Explanation:
For this exercise we use the law of conservation of energy
Starting point before touching the spring
Em₀ = K = ½ m v²
End Point with fully compressed spring
=
= ½ k x²
Emo = 
½ m v² = ½ k x²
x = √(m / k) v
x = √ (2.00 / 550) 10.0
x = 0.603 m
This is the maximum compression corresponding to the range of motion
A = 0.603 m
b) Let's write Newton's second law at the point of maximum compression
F = m a
k x = ma
a = k / m x
a = 550 / 2.00 0.603
a = 165.8 m / s²
With direction to the right (positive)
c) The value of the elastic force, let's calculate
F = k x
F = 550 0.603
F = 331.65 N
Answer:
Explanation:
Weight will be highest at the pole where there is no tangential velocity requiring centripetal acceleration. Also, due to the slight bulge of the earth at the equator, the distance from the surface to the center of mass of earth is slightly shorter there meaning gravity is slightly stronger. Fg = GmM/R²
Weight will be lowest at the equator where the object is moving about the earth axis at an angular velocity of 2π/(24(3600)) rad/s ( about 7.27e-5 rad/s)
This means that some of the (already weaker, see above) gravity there is required to supply the needed centripetal acceleration to keep the object on the ground.
Answer:
New force, 
Explanation:
The electrostatic force between two spheres is given by :

According to given condition, each of the spheres has lost half its initial charge, new force is given by :



So, the new force becomes (1/4)th of the initial force. Hence, the correct option is (d).