You have the exact right idea on how to solve the problem. You are going to want to start with basic kinematic equations for each car. Since the speeding car is moving at a constant speed, you get the equation

x_c = v*t

The police care starts at rest and accelerates, so you will want to use

x_p = a/2*t^2

Since no distances were specified, I am assuming that both cars start at the same point. If that is not the case you will have to add in the positions at t=0.

The point that the police catches the car is when their cars are at the same point (unless specified otherwise just ignore that cars aren't points) and you solve for t. You will have to be careful, the cars will be at the same point at two different times, so be careful in deciding which one.

Based on your wording regarding the magnitude, I am suspicious that you may have made a mistake in calculations. When working in multiple dimensions, you're going to have to be careful about how you get the magnitude.You have to treat each direction (x and y) separately and finding the total is not a simple average. Once you obtain the average velocity for each direction, the magnitude is obtained by

v = sqrt[(v_x)^2+(v_y)^2]

rather than averaging them. Essentially, you have to treat each component like a leg of a right triangle with the total as the hypotenuse.

Similarly, when it is time to get the direction you would calculate it like calculating the angle of a right triangle. Treat v_x as the adjacent side, v_y as the opposite side, and the magnitude v as the hypotenuse. Since you have all three of those you can pick your favorite trigonometric function to find the angle.

All magnetic materials can become magnetized because their magnetic dipoles align, but each type of material has a different mechanism for accomplishing this.

Paramagnets and diamagnets are much weaker than ferromagnets. Paramagnetism occurs when a material has unpaired electrons in its outer orbitals. Electrons are small magnetic dipoles because of their angular momentum in their orbitals. If you introduce an external magnetic field, these electrons will align with the field. Diamagnetism occurs because an external field will change the speed of electrons in orbitals. This change creates a magnetic dipole opposite the direction of the applied field.

Only a few materials are ferromagnetic, meaning that they can stay magnetized even if no external field is present. These are "permanent magnets" because the dipoles tend to stay aligned and make for a strong magnet.

When we talk about electric potential, we are actually talking about the potential difference between two points in space. In other words, a single value for the potential doesn't mean anything on its own; only the difference is meaningful to us!

We have to define where our potential is zero before we can talk about the potential at other locations. Typically, we define the potential to be zero infinitely far away. If this is the case, as it is for this problem, to find the potential everywhere else we have to integrate the electric field up starting from infinity. Fortunately, this has already been done for you in one of your formulas:

V = kQ/r

This formula is valid for a point charge, but can you also use it in this problem? Outside the symmetric spherical shell, the electric field looks the same as a point charge, so this formula is actually valid for r > R where R is the radius of the shell. Once you find the potential right outside of the shell, it's up to you to think about how the potential does or doesn't change as you go to the center of the sphere.

Power is the work done per unit time. In the case of a circuit, the electric power has to do with the charge carriers being forced to move across a potential difference. Imagine charge ΔQ moving across a potential difference V. You know that the potential energy difference is U = VΔQ, so the power is P = VΔQ/Δt. Written as a derivative we notice that the power can be written in terms of the current instead of charge.

P = V dQ/dt = VI

This is the basic formula for electric power. Whenever you have current flowing through a voltage difference, it took energy to make that happen. The power is the product of the current and the voltage.

The wave equation actually shows up in a lot of different contexts in physics as it is just a mathematical relationship. A few examples are a fluid like water or air, a string, and electric and magnetic fields. In each case, waves propagate through the medium.

In electromagnetism, you can derive the wave equation from Maxwell's equations. In a vacuum (meaning no charges or currents present), the equations in differential form reduce to:

∇·E = 0

∇·B = 0

∇×E = -dB/dt

∇×B = μ0 ɛ0 dE/dt

Taking the curl of the third equation, we get

∇x(∇×E) = -d(∇×B)/dt = - μ0 ɛ0 d^2E/dt^2

Using a vector identity for the left hand side, we find

∇x(∇×E) = ∇(∇·E) - ∇^2 E = - ∇^2 E

So we find that ∇^2 E = μ0 ɛ0 d^2E/dt^2, which is the wave equation where the wave speed is c = 1/sqrt(μ0 ɛ0).

This is just one example of how the wave equation is found in nature. It might be more detail than you wanted right now, but it should show you that whenever you can derive the wave equation in a medium, you will get waves that propagate through it! In this case, the waves are electromagnetic, like light for example.

Did you keep track of the units? rev/min is not in SI units, while N*m is. If you merely multiplied 678 by 4500, this would not work in Web Assign.

The electric field, by definition, points from high to low. Therefore, the field points toward the lower potential. Since the electron is negatively charged, the force on it is in the opposite direction of the field which is toward the high potential.

Hope this helps!

You need to express the effect of friction in terms of energy. The force of friction will be constant since the slide is straight. A constant force F applied over a single direction for a distance, d , does work on the object equal to F(dot)d. Now do the same conservatin of energy process, but with the work of friction added in.

Hope this helps!

From the energy conservation view, if the pulley was not massless, it would take >0 kinetic energy for it to rotate.

Hope this helps!