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...
[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...
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...
[SOLVED] Mininum Work to Pull Crate
Homework Statement
A student could either pull or push, at an angle of 30 degrees from the horizontal, a 50 kilo crate on a horizontal surface, where the coefficient of kinetic friction between the crate and the surface is 0.2. The crate is to be moved a...
Doc Al, you're wonderful. This was precisely the part I had trouble grasping.
I read your explanation and many times. Now I understand. I have to say that I found it sort of interesting that eq. 3 comes out to be the same if I assumed the opposite you did, that is, "right" is is negative and...
Doc Al, I'm sorry to make you explain this again but you keep stressing that the acceleration sign conventions are consistent I cant help but think that they really aren't. This might be a silly question, if I just multiple through by (-1) for eq. 1 then I am writing the acceleration as negative...
Doc Al, that explanation cleared it up for me I think.
So one direction is negative and the other is positive. I can ascribe a negative sign to the entire movement involving m1 moving down, m2 moving up, and m3 moving to the left, and vice versa. In other words, essentially there are only two...
I see that your equation was just the result of summing the equations 1,2, and 3. But how were you able to come up with it directly?
As I understand, the left side (m1+m2+m3)*a is the acceleration of the entire system of the 3 blocks. For the left side of the equation, what about m3*g? Did you...
(1) m1*g-T1 = m1*a
(2) T2-m2*g = m2*a
(3) T1-T2 = m3*a
The tensions are not in the same direction. I suppose because, I assumed +a applied to block m2 then +a must apply to m3 also. Although I have to admit I'm still not quite clear about this concept because the direction of acceleration of...
(1) m1*g-T1 = m1*a
(2) m2*g-T2 = m2*a
(3) m3*g-(T1+T2) = m3*a
About the sign convention for acceleration, I thought I needed to change the sign because as block m1 is accelerating say, upward, then m2 must be accelerating downward?
Okay. So let the left and right lines have tensions T1 and T2 respectively.
(1) m1*g-T1 = m1*a
(2) T2-m2*g = m2*a
(3) m3-(T1+T2) = m3*a
Would the above be correct?
[SOLVED] Acceleration of Block on an Apparatus
Homework Statement
See the attachment.
Homework Equations
The Attempt at a Solution
I am confused about whether the tensions of the two (left and right) parts of the line are the same or not. Since the three blocks are...
[SOLVED] Tension in a Line
Homework Statement
A rope is fixed at both ends on two trees, and a bag is hung in the middle of the rope, causing the rope to sag vertically. If the tree separation is 10 m, the mass of the bag is 5.0 kg, and the sag is 0.2 m, what is the tension in the line...
Homework Statement
A package is to be dropped from an airplane so that it hits the ground at a designated spot near some campers. The airplane, moving horizontally at a constant velocity of 140 km/h, approaches the spot at an altitude of .500 km above level ground. Having the designated point...