To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s
Answer:
True
Explanation:
This can be explained by the special theory of relativity for length contraction.
Length contraction is observed in the direction of motion of an object when an object moves with speed closer to the speed of light.
The length of the rocket in this case, appears shorter to the observer on earth in the stationary reference frame which is improper frame whereas the traveler in the rocket is in the same inertial frame which is proper for the rocket's size measurement.
Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
This is the unit of the energy
Answer:
1.73 seconds
Explanation:
The velocity the ball first hits the ground with is:
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)
v = -20 m/s
The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.
The time it takes to return to the ground is:
Δx = v₀ t + ½ at²
0 = (15 m/s) t + ½ (-10 m/s²) t²
0 = t (15 − 5t²)
t = √3
t ≈ 1.73 seconds
