Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration, 
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,



Frictional force is given by :


F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:

Two triangles that I marked are similar and from this we get:

The image and the object must have the same height so we get:

This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide. Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after. Subtract this from the the system total and you will have the momentum of the other object after the collision. Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.
Be careful with the velocities. They are vectors, so direction matters. Typically moving to the right is positive (+) and moving to the left is negative (-). It is not clear from your question which direction the objects are moving before and after the collision.
Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>
<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>
<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>
<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>