Recent content by azukibean

Thanks! That made a lot of sense.

1. Homework Statement The 4 black bars have constant linear density. In terms of L, find the coordinate of the center of massHomework Equations M0 X = summation of (MiXi) The Attempt at a Solution I understand most of the solution, except the answer key says one thing I don't get M/sqrt(2) at...

Thanks, you're awesome! The algebra was a total pain.

This is possibly one of the ugliest equations I've seen. I can't set the two Vf1 equal to each other, that would lead to 0. But I also can't figure out where to go from there.

Homework Statement A steel ball of mass m is fastened to a cord of length L and released when the cord is horizontal. At the bottom of the path, the ball strikes a hard plastic block of mass M = 4m at rest on a frictionless surface. The collision is elastic. Find the speed of the block...

Ah, that's it. Thank you so much.

I added the units to the original problem. There was no unit conversion involved, so I didn't want to clutter my work.

The first one is correct. The heat added or subtracted is equal to the change in internal energy and work. I'm unsure how to work the problem to find dQ; in ΔQ1→2 I equated it to Q1→2 because I assumed at "step 0" there was no heat energy. There are no omissions. V3 should actual be 6...

Homework Statement n = 5 moles, monatomic, ideal gas p1 = 7280 Pa; v1 = 2 m3 p2 = 7280; v2 = 6 p3 = 6500; v3 = 6 What is the change in heat, from 2 to 3, and from 3 to 1? Homework Equations PV = nRT U = (3/2)nRT Q = du + dw The Attempt at a Solution PV = nRT (7280)(2) =...

Cepheid variable! --- Straight legs: v0^2 = 2*-10*1.1 v0 = 4.69 Bent: 4.69^2 - 0^2 = 2*.5*a a = 21.996 60*22 = F(jump) - 600 F(jump) = 1920 N --- Thanks!

Homework Statement The jumper is 60 kg and wants to beat the school's record jump, 1.1 m. She bends her knees a distance of .5 m before jumping. How many Newtons must Fjump be to for her to beat the record? Homework Equations ma = ƩF V2 - V02 = 2a(X - X0) My teacher stated I would just need...

Thanks! Using the same equations as I did in part a I got 5250 m -> 5.25 km; a = -5/756; v0 = 50/6. Part B I used 250,000,000 for the mass instead of 1000 kg and got the right answer. This is kind of embarrassing. I must've stared at the problem, confused, for quite a while.

Dynamics -- supertanker coasting to a stop 1. Problem Assuming constant negative acceleration, how far does a supertanker of a mass of 250000 metric tons travel while coasting to a stop from an initial speed of 30 km/h in 21 min? What force is supplied by fluid friction of the water while the...

IIRC, the answer yields around 58 m; 8 m to spare. Intuitively, you can see that the range of X is greater than X when Y = 3 so it's not getting closer to 1.12 m. 0t^2 + 20t + 0 = 50 t=2.5 -5(2.5)^2 + (14.15)(2.5) + 0 = S S = 4.125 m; excess of 1.125 m Thanks!

I think it is asking me, length-wise, how much I have to spare, not height-wise; it asks me "is it long enough"? What makes you say the answer is suspicious?