Answer:
Speed of 1.83 m/s and 6.83 m/s
Explanation:
From the principle of conservation of momentum
where m is the mass, 
is the initial speed before impact, 
and 
are velocity of the impacting object after collision and velocity after impact of the originally constant object
Therefore 
After collision, kinetic energy doubles hence
Substituting 5 m/s for then
Also, it’s known that hence 
Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then 
Substituting, 
Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= 
replacing ;
Δx = 2t ; we have:
2t =
Given that thickness t = 700 nm
Then
2× 700 = --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness 
∴ 
Thus, the smallest thickness t of the air film = 140 nm
The given question is incomplete. The complete question is as follows.
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 
, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
Explanation:
We will calculate the work done as follows.
W = 
= 
= ![[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}](https://tex.z-dn.net/?f=%5B14000x%20%2B%205000x%5E%7B2%7D%20-%208666.7x%5E%7B3%7D%5D%5E%7B0.54%7D_%7B0%7D)
= 7560 + 1458 - 1364.69
= 7653.31 J
or, = 7.65 kJ (as 1 kJ = 1000 J)
Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.
Answer:
The velocity of the truck after the collision is 20.93 m/s
Explanation:
It is given that,
Mass of car, m₁ = 1200 kg
Initial velocity of the car, 
Mass of truck, m₂ = 9000 kg
Initial velocity of the truck, 
After the collision, velocity of the car, 
Let 
is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.
So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.
