Recent content by rlmurra2

yeah, that's the one I am using...wow I really don't think this problem is that hard. even tried it with period in minutes and seconds. oh well thanks anyways

I've been using Newton's third law all along, and its not working. I'll try again. What is it, just to make sure I am using the right formula?

Yep, that's what it says in the problem. Are we supposed to use Kepler's third law or whatever to solve this? It seems really easy that way, but when you plug in the numbers, it just doesn't work out.

I tried converting it to seconds and leaving it in minutes...still the answer is way off.

A satellite circles planet Zeron every 98 min. The mass of the planet is known to be 5.0 x 10^4 kg. What is the radius of the orbit? I don't know what I'm doing wrong, I am using the formula T^2=(4(pi)^2/GM)r^3...and I am not coming up w/ the right answer, I have like 5 choices to choose...

What is the mass of the Earth's atmosphere? The radius of the Earth is 6.4E6m. The only thing I can think of is to subtract something from the mass of the entire Earth or something...

An object is thrown downward w/ initial speed (t=0) of 10 m/s from height 60 m above the ground. At the same instant (t=0) a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other? Ok ... I...

Suppose the Sun's gravity were turned off, and the planets would leave their nearly circular orbits and fly away in straight lines described by Newton's first law. Would Mercury ever be farther away from the Sun than Pluto? If so, how long would it take for Mercury to achieve this passage...

Tcos(x)-u(mg-Tsin(x)=0 Tcos(x)-umg+uTsin(x)=0 Tcos(x)+uTsin(x)=umg T(cos(x)+usin(x))=umg T=umg/(cos(x)+usin(x) T=(.350)(1.30kg)(9.80m/s^2)/cos(x)+.350sin(x) This is where I am at, I understand the part that comes next, where it comes from.

Ok I am getting this now...I just don't understand at the part where the numerator has to be 0, where does the \frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)} {(cos\theta + \mu sin\theta)^2} [B]For this to be zero, the numerator must be zero, -sin\theta+\mu cos\theta = 0...

Sorry I posted that after I saw your post.

Fx=Tcos(x)-fs=max Tcos(x)=max+fs -> T=max+fs/cos(x) Fy=n-mg=may n=may-mg = m(ay - g) ?

I feel like I'm going in a big circle here.

Ok so T\cos \theta - F_{s} - mg = ma...but would be normal force be relevant?? I still don't know how to find the angle, without knowing the tension..

This one's not too hard...A 1.30kg toaster is not plugged in. The coeffficient of static friction b/t the toaster and horizontal countertop is .350. To make the toaster start moving, you carelessly pull on its electric cord. a) for the cord tension to be as small as possible, you should pull...