Answer:
Tarzan will be moving at 7.4 m/s.
Explanation:
From the question given above, the following data were obtained:
Height (h) of cliff = 2.8 m
Initial velocity (u) = 0 m/s
Final velocity (v) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 2.8)
v² = 0 + 54.88
v² = 54.88
Take the square root of both side
v = √54.88
v = 7.4 m/s
Therefore, Tarzan will be moving at 7.4 m/s at the bottom.
Well, if the bag is moving across the floor at a constant velocity. Then I would assume that the Fnet force is equal to 0, and thus the value of acceleration would also be equal to 0.
Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
Answer:
Ok I'm not 100% on this one but, try 3 lifes sorry if u get it wrong D:
Explanation: