Recent content by JWDavid

thank you, You are correct I did forget to square the angular speed. But on the average power - it does work both ways. If you note I wasn't taking the work done I was taking the change in Power/2 {(torque*angular speed)-0}/2. thanks again for your observations, I would probably have lost...

Thanks in advance - it went smoothly so I'm hoping I'm right but... usually that's when I've made a big mistake. Homework Statement An airplane propeller is 2.08 m (tip to tip) and has a mass of 117 kg. When the engine is first started it applies a constant torque of 1950 N m to the...

See NOW THIS is the problem with growing up not being allowed to use calculators! I did this portion in my head - and got it way wrong! Thanks got it now.

you're right I did forget g, and I did forget to transfer the error correction I noted from my paper of (DECIMAL) 854 not 8.54. But I'm still not feeling it: I now have the \Deltap = -.854Sin27.4 - .854SIN27.4 = .686 Ns for the change in impulse. I now have the time (calculated using...

Okay, so t should be 2*root(...) or 1.52 meaning that if it is possible to prove c) then 1.52s * 0.0524g (ball mass) should equal delta-p... but this means that delta-p only equals 0.0796 which can't be right?!? Because if I vector calculate p2-p1 for p(x) = 0 for p(y) = -8.54SIN27.4 -...

Homework Statement A 52.4 g ball is thrown from the ground into the air with an initial speed of 16.3 m/s at an angle 27.4o above the horizontal a) What are the values of kinetic energy of the ball initially and just before it hits the ground b) Find the corresponding values of the momentum...

So I think I've gotten it: Here's what I did please confirm. cm1 = (10.8*0 + 46.4*(L/2))/57.2 = .4L cm2 = (10.8*8.5 + 46.4*(L/2))/57.2 = .4L + 1.6 cm2 - cm1 = 1.6 which ..? means the boat moves under the dog 1.6 feet so the final answer is 21.4 - 8.5 + 1.6 = 14.5 feet - is this correct?

I believe you are forgetting the turn ratio

Homework Statement A dog weighing 10.8 lb, is standing on a flatboat so that he is 21.4 ft from shore. He walks 8.5 ft on the boat toward shore and then halts. The boat weighs 46.4 lb, and one can assume there is no friction between it and the water. How far is he from shore at the end of...

try this equation y = F(x): = yo[SUB] +xTAN\theta - gx2/2V[SUB]o2COS2\theta

thanks

Homework Statement Okay left my book at home - foolishly thinking I had all my equations - oops Two Skaters, one with mass 65kg and the other with mass 42kg stand on an ice rink holding a pole with a length of 9.7m and a mass that is negligible. Starting from the ends of the pole the...

Is NO ONE able to provide help with this?!?

No. t = sqrt(20*2/9.8). One thing I've found in the few months I've been doing physics, you have got to reality check you answers (ALMOST every time). 20m =~ 60 feet it doesn't take even a feather 15 seconds to fall 60 feet. The answer for time is going to be somewhere around 2 seconds...

still need help on this one