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
The point in which it originates.
To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.
This definition is described in the following equation as,

Where,
permeability of free space
Number of turns in solenoid 1
Number of turns in solenoid 2
Cross sectional area of solenoid
l = Length of the solenoid
Part A )
Our values are given as,





Substituting,



PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.
How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.
Hope this helps.
Answer:
Explanation:
Mass of ball Is m=96.1g=0.0961kg
Height above spring is 59.1cm
L=0.591m
Extension of the spring is 4.75403cm
e=0.0475403m
Then the distance the ball traveled is H=L+e
H=0.591+0.0475403
H=0.6385403m
Then, the potential energy of the ball is given as
P.E=mgh
P.E=0.0961×9.81×0.6385403
P.E=0.602J
From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another
Then, the P.E is transferred to the work done by the spring
Then, Work done by spring is given as
W=½ke²
W=P.E=½×k×0.0475403²
0.602=½×k×0.0475403²
k=0.602×2/0.0475403²
k=532.72N/m
The spring constant is 532.72 N/m