Recent content by microguy

Did you mean Mean or Median?

No, it did not say both changed, it asked you to recall which one(s) changed? Thought experiment: Take a water-proof speaker generating a 440 Hz sine wave to the edge of a swimming pool. Now submerge yourself with the speaker in a swimming pool. What difference does your ear perceive?

Whew! I'm not trying to beat this problem to death, but it was beating me to death, so I had to pursue it. I re-read the Gravity chapter in my old textbook and realize--no--I can't approach the cylinder as a point mass/sphere (although the spherical result is rather interesting). The results...

Not trying to be bull-headed but I am feeling lost here now... (a) How does density change when M/V = (M/2)/(V/2) (b) OK now you say density does not change. Huh? But this I agree with! But what I do not follow is "They [sic] guy remains the same distance as before from the part of the mass...

I am quite flummoxed by this problem. Quite possibly, I do not recall where the gravitational force equation is supposed to work. Also, where are you going with the density argument? The density of the planet is not changed (?), only its mass and size. On one hand, indeed, it doesn't make sense...

Forum, please correct me where I am wrong in my thinking. Forget about cylinders, spheres, slices, and asteroids. If I am distance D1 from point mass M1 I weigh F1. If I am distance D2 from point mass M2 I weigh F2. The ratio F2/F1 invokes only D1, M1, D2 and M2. We can treat the "cylinder"...

@ nrqed: But you are leaving out half the puzzle, which is the person is half the distance from half the mass. If half the Mass only: f/F = 1/2 If half the d only: f/F = 4 If both: ...

Good hint PhantomJay. I was trying to solve the OP's question as well and got hung until I read this.

Step back a bit. There are two scenarios, 1 on Earth) y(1) = v0(1) X t(1) + 9.8 X g X t(1)^2 2 on Moon) y(2) = v0(2) X t(2) + (9.8/6) X g X t(2)^2 Now, what do you know about y(1), y(2), v0(1), and v0(2)?

Sorry I forgot to ask also -- where are you getting 14? That was the initial velocity of ball 1. The initial velocity of ball 2 is unknown -- it's what you're supposed to be solving for (^: The distance traveled is known--or can be determined by plugging ball 1 into d = 1/2 (v_initial -...

Where are you getting 0.7? The problem states "half a second later", not "half the time later".

Hmm. I disagree. Think of the extremes: If the rope is very long such that the cable is nearly horizontal, it's almost as if the train and carriage were on the same plane. Disregard the weight of the cable of course (^: On the other hand, if the rope is only an inch longer than the height...

Thanks for that! This makes sense and you helped me notice something I was not doing quite right.

Modelling "Gravitational" Motion Hi All, I am new here and this is not a homework question... I'm turning 40 this year and my retention of physics is low to say the least. I am trying to rekindle some knowledge... Just for fun! I am trying to use a spreadsheet to model the motion of a...