Answer:
4.2 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v
315 g m/s − 525 g m/s = -525 g m/s + (75 g) v
315 g m/s = (75 g) v
v = 4.2 m/s
Answer:
15 m/s
Explanation:
Using the law of conservation of energy, potential energy equals kinetic energy hence
Therefore
where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet
Taking g as 9.81 then
Given that, A basesball is dropped from 100 meters above the surface of the earth. If the same baseball was dropped from 100 meters above the surface of the moon, it will take more time to hit the ground as compare to the ball dropped on earth. This is moon's gravity is one-sixth of that of earth. Object falling on earth possess more force of attraction. So it will reach the earth in lesser time. At moon the force of attraction is low compare to that of earth. Object will take more time to reach the surface.
D
The motion of an object in a circle at a constant speed.
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).
</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>
</span>
</span>
