Answer:
Q = c M ΔT where c is the heat capacity and M the mass present
Q2 / Q1 = M2 / M1 since the other factors are the same
M = ρ V where ρ is the density
M = ρ Π (d / 2)^2 where d is the diameter of the sphere
M2 / M1 = (2 D/2)^2 / (D/2)^2 = 4
It will take 4Q heat to heat the second sphere
200J
Explanation:
K.E=½mv²
K.E=½×16kg×5m/s²
K.E=8kg×25
K.E=200J
Tect book
potential energy = mass × gravity × height
so, change in potential energy = mass × gravity × change in height
2 = 50 × 10 × Δh
2 ÷ 500 = Δh
Δh = 0.004 m
This distance does depend on the initial velocity of the ball.
