Answer:
R=2F
Explanation:
As the forces are in same direction so the resultant force will be:
R=F+F
R=2F
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:

where

is the coefficient of friction, while N is the normal force. So we have:

since we know that F=300 N and

, we can find N, the magnitude of the normal force:

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is

due to the presence of the oil. Therefore, we have:
Mass of yellow train, my = 100 kg
Initial Velocity of yellow train, = 8 m/s
mass of orange train = 200 kg
Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)
To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which
The sum of initial momentum = the sum of final momentum
Since the question only wants the sum of initial momentum,
(100)(8) + (200)(-1) = 600 m/s
Answer:
14.8 m
Explanation:
S= ut +
a
where u = initial velocity
S= (0
)(2
) +
(7.4
)(2
)
S=
(7.4
)(2
)
S=14.8 m
Explanation:
It is given that,
Wavelength of red laser light, 
The second order fringe is formed at an angle of, 
For diffraction grating,

, n = 2


The wavelength λ of light that creates a first-order fringe at 22 is given by :




Hence, this is the required solution.