Oh ok! So I redid the problem and I'm left with M_{m}*\frac{1}{rm^2}/M_{e}*\frac{1}{re^2} (where rm is Gerald's distance from the moon). I'm assuming that's my real answer.
Sure. I have:
Ratio = Fm/Fe
Thus:
Ratio = G*M_{M}*M_{E}*\frac{1}{rm^2}/G*M_{M}*M_{E}*\frac{1}{re^2}.
So the G and M's cancel out and I'm left with \frac{1}{rm^2}/\frac{1}{re^2}
That equals \frac{re^2}{rm^2}
Where rm = Radius of Gerald from the moon
and re = Radius of the Earth...
After taking another stab at the problem, I got another wrong answer. What I was trying to do was find Fm and Fe (through the formula posted). I used the distance that Gerald was from the moon (I used that as the radius value) to find Fm, then I found Fe and divided Fm/Fe. Could you explain to...
Well, now that you mention it, it seems like I was finding the ratio of attraction between the Earth and the moon...
Those values were given, so I'm assuming that's what I'm supposed to be using. Alright, I think I get this. I shouldn't need to worry about Gerald's mass right?
Homework Statement
Gerald stands on the roof looking up at the full moon, which has a mass of 7.39 x 10^22 kg. At this moment, Gerald is 3.77 x 10^8 m from the center of the moon. If the mass of the Earth is 5.96 x 10^24 kg, and the radius of the Earth is 6.39 x 10^6 m, what is the ratio of...
Yeah I figured that \muk was the right one to use. I went through with it and found the right answer, Just that and a couple sign errors got me. Also for the last two parts, they're literally just asking the same thing. When, ax = 0, what is F, while part one is saying if F = 0, find ax. I could...
Oh wow, I didn't even notice that! Alright, so friction is moving against the crate moving downhill, which means that it's not -f, but +f.
As for the next two parts, wouldn't a constant speed mean that a = 0? Specifically ax. So if that's right The F for uphill would be F = MgSinθ + \mus*N?
Homework Statement
A crate is being pushed up a hill with friction. Given: m1 = 131 kg, θ = 23°, \mus = 0.15, and \muk = 0.08. If the crate isn't being pushed anymore, what will be the magnitude of the acceleration of the crate sliding downhill?
What force must be exerted to move the crate...
Alright so B is 4km but I'm not really sure about the direction, is it the θB = 328°? I'm sorry if I'm missing something, I have a hard time understanding vectors.
EDIT- I found the coordinates! It turned out that y=14.36 km. Before that I did some math to find the x and y for B, which ended up...
Alright, so the angle is already given to me is 328°. So does that mean that when I'm drawing my vectors, the angle between A and B is that degree since θB = 328°? Furthermore C should be a point, not a line, right? It's 4A-3B (in relation to the two vectors).
Homework Statement
On a treasure map,
A = -5 (km)x + 2 (km)y, B = 4 km, and theta = 328 deg. The treasure is located at C = 4A - 3B. What is the x-coordinate of the treasure?
What is the y-coordinate of the treasure?
Homework Equations
a^2 + b^2 = c^2
Vector addition
The...
Homework Statement
(Yes this is a question about Batman.)
To escape from Batman, 4 villains on motorcycles drive
horizontally off the edge of a parking garage, landing in a pile
of sand. The riders all have different initial speeds: v0A =
14m/s, v0B = 24m/s, v0C = 18m/s, and v0D = 19m/s...