Formula for potential energy is V=mgh, where m is mass in KG, g is earth acceleration (10 m/s^2), and h its height in meters. We know mass, acceleration is constant and also known, we know height also. Lets substitute
V=75*10*300=225000[J]=225[kJ] - its the answer
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
(B) The speed is larger at A than at B.
Explanation:
Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.
