Answer:
Δ T = 2.28°C
Explanation:
given,
mass of marble = 100 Kg
height of fall = 200 m
acceleration due to gravity = 9.8 m/s²
C_marble = 860 J/(kg °C)
using conservation of energy
Potential energy = heat energy
Δ T = 2.28°C
Answer:
t1 = t2 + 3.02 V = 41.5
V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2
Both stones reach the same height after the specified times
V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)
2 V / g = t1 + t2 = 2t1 + 3.02
t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s
t2 = t1 + 3.02 = 5.74 sec
Check:
41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m
41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m
Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s
