If the moon was hit by an asteroid there would be a crater mark and possible movement.
The answer is B (The second one). I'm not sure though.
Answer:
a. 
b. 
c. 
Explanation:
First, look at the picture to understand the problem before to solve it.
a. d1 = 1.1 mm
Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:
To solve the equations we have to convert all units to those of the international system. (mm→m)

μ0 is the constant of proportionality
μ0=4πX10^-7 N*s2/c^2
b. d2=3.6 mm
Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:
J: current density
c: outer radius
b: inner radius
The cilinder's current is negative, as it goes on opposite direction than the wire's current.




c. d3=7.4 mm
Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.
Explanation:
Given that,
Initial speed of the bus, u = 0
Acceleration of the bus, a = 0.5 m/s²
Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.
So,

Let d is the distance cover during that time. So,

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.
Answer:
The uncertainty in the location that must be tolerated is 
Explanation:
From the uncertainty Principle,
Δ
Δ

The momentum P
= (mass of electron)(speed of electron)
= 
= 
If the uncertainty is reduced to a 0.0010%, then momentum
= 
Thus the uncertainty in the position would be:
Δ
Δ