Answer:
B) shrinks
Explanation:
The magnetic force is a force exerted between two magnets, or two magnetic materials, or also on an electric charge moving in a magnetic field.
If we talk about magnetic material, the magnetic field they generates can be represented using a dipole: essentially, they have a north pole (where the lines of the field go out) and a south pole (where the lines of the field go in).
Also, the lines spread apart as we move away from the magnet itself. This means that the strength of the field (and so, the intensity of the force) decreases as we move away from the magnet.
Using this description, we can now understand that when we move the paper clip further from the magnet, the force exerted on the clip decreases, as the magnetic field becomes weaker. So, the correct answer is B.
I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is

And the time taken is t=22.4 s, so the average speed of the player is

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is

and so, the average velocity is
Waves in which the displacement of the medium is in the same directions as, or the opposite direction to, the traveling wave's direction.
A wave in which the particle movement is parallel to the wave's direction.
I gave to definition, I hope you understand it. Have a nice day :)
Answer:
block K = 29.39 J and spring #1 Ke = 360 J
Explanation:
In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work
= Ef - E₀
Let's look for the energies
Initial
E₀ = Ke = ½ k₁ x₁²
Final, this is just before starting to compress the spring
Ef = Ke = ½ m v²
The work of the rubbing force is
= -fr x
Let's write Newton's second law the y axis
N-W = 0
N = W
fr = μ N
fr = μ mg
Let's replace
-μ mg x = ½ m v² - ½ k₁ x₁²
v² = 2/m (½ k₁ x1₁² -μ mg x)
v² = 2/6 (½ 2000 0.6²2 - 0.5 6 9.8 1) = 1/3 (360 - 29.4)
v = 3.13 m / s
With this value we calculate the energy of the block
K = ½ m v²
K = ½ 6 3.13²
K = 29.39 J
Calculate eenrgy of the spring ke 1
Ke = ½ k₁ x₁²
Ke = ½ 2000 0.60²
Ke = 360 J