Answer:
Explanation:
Given that
Mass , m = 25 kg
We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows
Static friction force ,fs= 165 N
Kinetic friction force ,fk = 127 N
If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.
Lets take coefficient of kinetic friction = μk
The kinetic friction is given as follows
fk = μk m g
Now by putting the values
127 = μk x 25 x 9.81


Therefore the value of coefficient of kinetic friction will be 0.51
Answer:
<em>The rubber band will be stretched 0.02 m.</em>
<em>The work done in stretching is 0.11 J.</em>
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = <em>0.02 m</em>
<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.
potential energy stored = 
==>
= <em>0.11 J</em>
Answer:

Explanation:
The speed of light is given by
and
hence

Speed of light is given by

Answer:
The width of the central bright fringe is 7.24 mm.
Explanation:
Given that,
Wavelength = 632.8 nm
Width d= 0.350 mm
Distance between screen and slit D= 2.00 m
We need to calculate the distance
Using formula of distance

Put the value into the formula


We need to calculate the width of the central bright fringe
Using formula of width

Put the value into the formula


Hence, The width of the central bright fringe is 7.24 mm.
Answer:
The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
Explanation:
Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.
Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.
Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.