Recent content by aligass2004

r is the radius of the earth. I still just don't understand what it's asking for.

The orbital period is 24 hours or 86400 seconds. I think the relevant equation is v = 2(pi)(r)/T.

That was the problem. Thank you! http://i241.photobucket.com/albums/ff4/alg5045/1012540D.jpg Now it asks for the value of the composite constant (GmE/r^2), to be multiplied by the mass of the object mO in the following equation? http://i241.photobucket.com/albums/ff4/alg5045/render.gif

Homework Statement http://i241.photobucket.com/albums/ff4/alg5045/MUG_co_6.jpg A satellite that goes around the Earth once every 24 hours is called a geosynchronous satellite is in an equatorial orbit, its position appears stationary with respect to a ground station, and it is known as a...

Homework Statement http://i241.photobucket.com/albums/ff4/alg5045/1012540B.jpg Consider the Earth following its nearly circular orbit (dashed curve) about the sun. The Earth has mass 5.98 *10^24 kg and the sun has mass 1.99 *10^30 kg. They are separated, center to center, by r=93 million...

Ok, but that still doesn't help me figure out the horizontal equation for m1.

I don't have a friction force on m1 because I'm not sure where it would go. Would it go to the right? Opposite the friction force for m2? And if that's right, I know the vertical equation is N = m1g and I'm not sure what the horizontal equation would be. I know for m2, it's moving to the...

Homework Statement http://i241.photobucket.com/albums/ff4/alg5045/MLD_2l_15_001.jpg A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box in us. The...

I figured it out. I don't know why I keep using the wrong angles. That's what was screwing me up I think. Thanks for your help.

Ok. I got everything for system b, and I thank you for your help with that. Now I need help with system a. Would the equation be to solve for A for system a: A[sin(150)cos(45) + cos(150)sin(45)] = 250cos(45)

The second equation would be B sin(45) = A cos(210) +250

I meant 685 for A for system b. My mistake.

Well I already found A for system b = 685N, which was correct. However, when I plugged 685 into the horizontal equation, I got B = 484N, which was incorrect. I've also tried solving for A for system a. For that I got 685N again, but it wasn't right. I know how to solve them. We've done...

For system a, the horizontal equation is Ax = Bx or Acos(thetaA) = Bcos(thetaB), and the vertical equation is Ay + By = C or Asin(thetaA) + Bsin(thetab) = C. For system b, the horizontal equation is again Ax = Bx or Asin(thetaA) = Bcos(thetaB) because A is below the horizon in this case. The...

Homework Statement http://i241.photobucket.com/albums/ff4/alg5045/image546.jpg Find the tension in each cord in the figure if the weight of the suspended object is 250N. Part A) Find the tension in the cord A for system (a) Part B) Find the tension in the cord B for system (a) Part C)...