The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value
Answer:
1) 51 m
2) Some energy was transformed to other forms. (This question)
3) 3.24 J
4) 45 J
5) 1020 J
Explanation:
100%
I would answer but you have no choices lol
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) =
=
= 392 J
(C) average frictional force = 
- change in KE (ΔKE) = initial KE - final KE
- ΔKE =
-
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] =
=
= 212.33 N