Answer:
B.
It will be greater than 10 J.
Explanation:
The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):
E = PE + KE
According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.
The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.
The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.
Here at the beginning, when it is at the top of the roof, the baseball has:
PE = 120 J
KE = 10 J
So the total energy is
E = 120 + 10 = 130 J
As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).
Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.
More energy is released in nuclear reactions than in chemical reactions; this is because in nuclear reactions, mass is converted to energy. Nuclear energy released in nuclear fission and fusion is several 100 million times as large as an ordinary chemical reaction like the combustion process. The reason why nuclear energy release so much energy is because tremendous amounts of energy is released at one time. The nuclei in a nuclear reaction undergo a chain reaction, causing the neutrons to move extremely fast and release high amounts of energy.
There are no appropriate units for power on the list you provided
Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²