A 5.00-kg block is in contact on its right side with a 2.00-kg block. Both blocks rest on a horizontal frictionless surface. The 5.00-kg block is being pushed on its left side by a horizontal 20.0-N force. What is the magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block?

Explanation:

The relation between force, mass and acceleration is as follows.

               F = ma

or,          a =

As two blocks are in contact with each other. Hence, total mass will be as follows.

                mass = 5 kg + 2 kg

                          = 7 kg

Now, we will calculate the acceleration as follows.

             a =

                =

Hence, force exerted by mass of 2 kg on a mass of 5 kg will be calculated as follows.

                    = 5.714 N

Thus, we can conclude that magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block is 5.714 N.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.