Question
What was the initial momentum of the bullet before collision?
Answer:
10 Kg.m/s
Explanation:
Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s
Answer:
=24.25 ^−1
Explanation:
Let and be initial and final velocity of the body respectively,
be acceleration due to gravity ( 9.8^−2 ), ℎ be the height of the body.
=0 ^ −1
ℎ=30
we know that, ^2−^ 2=2ℎ
^2=2∗9.8∗30
^2=588
=24.25 ^−1
Answer:
C. both forces have the same magnitude
Explanation:
Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.
Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.
We have the equation for momentum as:

Newton's second law is Mathematically given as:

Momentum is constant and the reaction time is equal, so the force exerted will also be equal.
Each principal energy level has one sublevel containing one orbital, an s orbital, that can contain a maximum of two electrons. Electrons in this orbital are called s electrons and have the lowest energy of any electrons in that principal energy level.