Recent content by mrlucky0

Oh good point, I didn't think of that! You're saying that there is only one initial velocity for the bottom-up rock to be thrown such that it just reaches the top of the building.

Oops, I meant to say, they assumed they meet halfway through their flight time. The result here is that for any height, the thrown rock always travels 3x more than the dropped rock, so in this case 75m and 25m respectively. Does that sound right? Anyway, it's just one of the "combinations,"...

Yes. In looking at the solution, they assumed that the rocks meet exactly halfway through their flight. Interestingly this is also the time when the thrown rock reaches it's max height.

One man drops a rock from a 100 m building. At exactly the same moment, a second man throws a rock from the bottom of the building to the top of the building. At what height do the rocks meet? I have trouble understanding how there can even be an answer to this problem without more info...

Homework Statement Object A and B are placed on a spring. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into air, object A will: answer: rise to the same height as object B Homework Equations The Attempt at a...

Thanks, I've got it now. Somehow I completely overlooked the fact that, friction neglected, the system would be conservative. I mean, from a "practical" standpoint from which I based my thought process on when I encountered the question, it's just hard to believe any sane person would want to...

A girl riding her bicycle up a steep hill decides to save energy by zigzagging rather than riding straight up. Ignoring friction, her strategy will: A. require the same amount of energy but less force on the pedals B. C. D. I missed this question and when I saw the correct answer, I couldn't...

Thanks. Your tip helped me solve the problem!

[SOLVED] MCAT Passage: Inelastic Collision Homework Statement A car (1000 kg) and a truck (2000kg) start from rest on a long, straight track. At time t=0, the truck is at position x=0 and the car is at position x=100m. Both vehicles then accelerate toward each other and collide. Both the...

Thanks, I've got it now.

I'm working on a chemistry problem and I'm trying to follow the derivation for % ionization (of a weak acid or base) but I can't seem to understand how rearrangement of eq. 6 results in eq. 7. In more general terms, I started with the Hendreson Hasselbalch equation: pH = pka + log(A-/AH)...

You need to find the force friction of the ladder (with the human) with respect to the ground. This is given by uk*N. Where uk is the coefficient of friction and N is the normal force.

The more I reason it, the more I am convinced it's correct. The initial work that I found pertains to the minimum work required when the crate is pulled at a 30 degree angle. Of course, the absolute minimum work would involve applying a horizontal force. Hence it's F*cos(30). Thanks blochwave...

Following your directions: uk*(mg-101 N*sin(30))*d = 1318 J This is the same number as 101 N*cos(30)*d so the two cancel out don't they? Edit: I think you're absolutely right about taking horizontal component of the force I got since the problem is asking for the minimum work. Ie. it's...

The solution in my textbook is ~1200 N, so that has me second-doubting myself.