**By decree of His Brilliance, we provide you with the necessary understanding to complete the webassign homework:**

__Circular Motion:__

1). This question is pretty cancerous. The basic premise is that the lower the radius the greater the acceleration. If you don't know the radius of the half-circles then you should drop His esteemed academic, but the quarter-circles are pretty hard; you've got to find the length of the line which encompasses the circumference and the point where the curve becomes a line.

2). Necessary equation here is: v = (2πr/T)

T is in seconds (obviously) and r is the earth's radius **PLUS **the orbit height above Earth.

The Centripetal acceleration is just v^2/r (and if you still don't understand, you deserve to fail).

5). This question wasn't that pleasant to taste, but it's easy once you understand that the time period is 24 hours (i.e. it takes one day for the earth to rotate around its axis (wow!)).

The second part is easy, **BUT WATCH OUT! **The answer wants it in **Minutes**. Because what better way to throw someone off by asking for the answer in an irrelevant unit: bastards.

6). This one is easy. The wording about the limited acceleration may 'throw' (get it) some people off, but it's really just saying this is the lowest acceleration it can be, what's the highest radius?

Again, it's v^2/a that we have to use.

The second part is easier, and if you don't get it then ...well let me say, 'Switzerland exist for a reason, folks!'.

7). Okay, this is **literally **the only challenging part of the assignment. It's a good challenge, though, unlike some of the others.

Basically, one could have probably deduced, by simple understanding, that we're looking at a kinematics question with a horizontal and vertical component, i.e. projectiles.

Now, this question has three parts to it: find the time in the vertical component, find the initial/final velocity in the horizontal component (acceleration is 0 in the horizontal), find the centripetal acceleration by subbing in your initial velocity as v in v^2/r.

If you can't do this, when I've literally revealed the procedure, then hand your head into Chris because you're clearly not worthy to be in his tutelage.