Recent content by matxjos

Okay...after more thought, I am convinced my solutions manual is wrong. I hate how many errors are in these things. Thanks jedishrfu.

I feel so dumb. I cannot figure out this relationship for the life of me.pH= -log[H+] Imagine 2 concentrations. If the hydronium ([H+]) concentration differs by a factor of 2, it means the pH of the 2 solutions differs by -log √2 HOW?! So basically, what they're saying is (logx)(log√2)=log2x...

Wow, thanks, I got the answer! I didn't realize things cancel out...physics is most confusing when that has to happen. But very interesting. Thanks haruspex and gneill! :biggrin:

Wait, if there is no rotation after the ball leaves the table, does that mean the rotational energy is zero? Or just the same as the initial KE rot?

Well, a solid sphere's moment of inertia is 2/5MR^2. Is this the expression I use? I still don't know the radius... I'm not even sure what you mean to write an expression for the rotation rate. Like in rad/s? I know how to get that from rev/min, or even m/s, but I don't know the radius...

Homework Statement A 0.125kg basketball is rolling w/out slipping on a horizontal table at 4.50 m/s when it rolls off the edge and it falls to the floor, 1.10 m below. What is the rotational kinetic energy of the ball right before hitting floor?Homework Equations KE rot: .5 I w^2 KE...

Nvm, I figured it out. Was using cosine instead of sine. Thanks for the help, questions done. :smile:

Okay. I assume you have to use geometry to figure that out. Tried to use cosine. Can you tell me how to figure out that distance?

Hi, thanks for the response. That is what I initially thought, but what I got isn't an answer choice. I calculated that the 1.2kg mass is 1.788854383m from the axis of rotation. so 1.2*1.78^2 = 3.84 kg*m^2 This isn't an answer choice the options are: 50, 19, 11, 0.96, 0.72, 29...

Homework Statement [refer to picture, thanks] The "L-shaped" figure rotates on the axis which which intercepts through 9 kilogram and 2.5 kilogram masses. Find the moment of inertia of the object for this type of rotation. Disregard the masses of connecting bars. Homework Equations MR^2...

oh ok i see now that the acceleration is 0...but how does this help me figure out when the trains meet and how long before the do?

phrak how did you know the acceleration terms dropped out? i used your equation though and i got t=45 - 20 m/s tain and t=36 - 25 m/s train but i left the distance 900 in km.

ummm so i used the v^2 = v(starting)^2 +2a(x-x[starting]) i got .222 = acceleration for the 20 m/s train and .34722222 for the 25 m/s train... now what?

ah shoot you made me realize i used velocity for the acceleration dang it. and yeah, i have no idea what time is supposed to be in. I'm guessing seconds. and also one has to change the 900 km to 900000 metres I'm pretty sure. i sort of have no clue on this problem...

ok i have x = x(starting distance) + v(starting vel.)t +.5at^2 v^2 = v(starting)^2 +2a(x-x[starting]) and v=v(Starting) +at i found the amount of time is 300 for the 20m/s train and 268.3281573 for the 25 m/s train. but how to found the time they meet?