a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio
1 answer:
The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
The given parameters;
- mass of the ball, m₁ = 0.8 kg
- speed of the ball, u₁ = 2.5 m/s
- mass of the object at rest, m₂ = 2.5 kg
- final velocity of the object at rest, v₂ = 1 m/s
Let the final velocity of the 0.8 kg ball immediately after collision = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)
2 = 2.5 + (0.8)v₁
-0.5 = (0.8)v₁
Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
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Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
Therefore, the number of turns of wire needed is 573.8 turns