Answer:
The kinetic energy when it returns to its original height is 100 J
Explanation:
The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J
Therefore the final height is given by
u² = v² -2·g·s
Where:
u = final velocity = 0
v = initial velocity
s = final height
Therefore v² = 2·g·s = 19.62·s
P.E = Potential Energy = m·g·s
Since v² = 2·g·s
Substituting the value of v² in the kinetic energy formula, we obtain
K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J
When the ball returns to the original height, we have
v² = u² + 2·g·s
Since u = 0 = initial velocity in this case we have
v² = 2·g·s and the Kinetic energy = 0.5·m·v²
Since m and s are the same then 0.5·m·v² = 100 J.
The volume of the sphere is given by V = 4/3 * pie * r^3. We seek dr/dt that the rate of change of the radius with respect to time.
V = 4/3 * pie * r^3
Since we know the rate at which the volume changes wrt time. We can plug it in to find dr/dt.
dV/dt = 36pie. But dV/dt = 4/3 * pie * 3r^2 * dr/dt
So we have that dr/dt = (dV/dt) / 4/3 * pie * 3r^2
So dr/dt = (36pie)/ 4/3 * pie * 3(4)^2
dr/dt = 113.112/ 201.0864 = 0.5624
The answer is the letter ( d )
Answer:
0.0316 m
Explanation:
Wok done = Energy change
Work done on the spring = Energy change of the block
Elastic Potential stored = Kinetic energy of the block
x = 
x = 
x = 0.0316 m
k = spring constant
m = mass of block
v = velocity of the block
x = compression of spring
Given:-
- Mass of the cart (m) = 35 kg
- Speed (consider Velocity) = 1.2 m/s
To Find: Momentum of the cart.
We know,
p = mv
where,
- p = Momentum,
- m = Mass &
- v = Velocity.
Thus,
p = (35 kg)(1.2 m/s)
→ p = 42 kg m/s (Ans.)
Conclusion:-
A. ☑️ 42 kilogram - metre per second.
