Recent content by kasiu

Oh! Thank you very much! The both of you have been very helpful! Once again, thank you! :)

So, quick question: in my final formula to find work of the kinetic would i set it up as Work = Force*cos(theata)*distance Work= 131.371*cos(23)*9 or would i use Work=131.371*cos(0)* 9 because cosine would be 0 degrees on the horizontal?

1. A 2.00 x10^2-N force is pulling an 75.0-kg refrigerator across a horizontal surface. The force acts at an angle of 23.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 9.00 m. Find (a) the work done by the pulling force, and(b) the...

ooooooh! I get it so, Ma-Mb =N* r^2/Gm So, what I posted before was right :) thank you so much! I understand it now! :):):) wow, that took me a while finally my brain made some connections thanks once again!

so acceleration= F/m or a=g=6.67e-11* mass/distance^2

G(Ma)(m)/d^2 - G(Mb)(m)/d^2= N* r^2/Gm Would I use that

do i get the acceleration?

I know that the equation has to equal (mass of planet A)-(mass of planet B)

so: (5370*acceleration due to gravity) + 3520 N = 5370 * gravity... that's not right I can see, but how do I find the weight. I can't solve for acceleration because I don't have the weight of the planet... I feel confused :(

How would I find weight? Would I use the equation attached?

So, would I set it up like this: W_subB + 3520 N = M * (g_subB) Or M * (g_subA) = W_subB +3520 N So, would I solve for the acceleration of planet A to get g_subA and then use 5370 Kg for my M (mass) and solve for W_subB

1. The mass of a robot is 5370 kg. This robot weighs 3520 N more on planet A than it does on planet B. Both planets have the same radius of 1.33 x 10^7 m. What is the difference MA - MB in the masses of these planets? 2. attached picture 3. I was wondering what to do with the...