There is too much information given, it's hard to understand exactly which variables are important in this problem.
Answer:
Explanation:
A 40kg child throw stone of 0.5kg
At a direction of 5m/s
Recoil can be calculated using recoil of a gun formula
m_1•v_1 + m_2•v_2
m_1•v_1 = -m_2•v_2
The negative sign show that the momentum of the boy is directed oppositely to that of the stone
m_1 Is mass of boy
v_1 is the recoil velocity of the boy
m_2 is mass of stone
v_2 is the velocity of stone
Then,
m_1•v_1 = -m_2•v_2
40•v_1 = -0.5 × 5
40•v_1 = -2.5
v_1 = -2.5 / 40
v_1 = -0.0625 m/s
The recoil velocity of the boy is 0.0625 m/s
Answer:
true blood season and answers the question of whether you are not the intended recipient you are not the intended recipient you are not the intended recipient
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
We are given an object that is speeding up on a level ground.
Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.
The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.
Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.
We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.
