Answer:
The K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Explanation:
Given data,
The mass of the bowling ball, m = 10 kg
The height of the building, h = 25 m
The total mechanical energy of the body is given by,
E = P.E + K.E
At height 'h' the P.E is maximum and the K.E is zero,
According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'
Therefore, P.E at 'h'
P.E = mgh
= 10 x 9.8 x 25
= 2450 J
Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Answer:
10^-7 C
Explanation:
m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0
Let q be the charge on bead
Force = m a = q E
a = q E / m
q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C
Answer:
A.
Explanation: both triple by 3
Explicacion
m = 65 kg
g = 10 m/s²
r = 0.5 cm (1m / 100 cm) = 0.05 m
A = π r² = π (0.05 m)² = 0.00785 m²
F =W = m g = 65 kg(10 m/s²) = 650 N
P = F/A = 650 N / 0.00785 m² = 82802.54 N/m²
