Recent content by postfan

Greg, do you know when the offer will be back?

Ok, so now what?

Ok so F is the component of the weight that is parallel to the incline, is that right?

F is the force produced from the power of the engine and the cos theta comes from the angle of the incline, the bigger the angle the less velocity.

I looked up the torque-velocity relation ,was I supposed to derive it somehow?

The force is the one that the car provides.

Force

P=F*v*cos(theta).

Homework Statement The maximum torque output from the engine of a new experimental car of mass m is τ . The maximum rotational speed of the engine is ω. The engine is designed to provide a constant power output P. The engine is connected to the wheels via a perfect transmission that can...

So this ? http://upload.wikimedia.org/math/e/5/6/e5631b9f8b287ee46ac6f3caf023b10c.png

Cool now what about 21?

Ok for 20 I did mg(L-Lcos(theta))=.5mv^2 and got v^2=2gL(1-cos(theta)). Substituting that into my "T" equation I got T=mg(3-2cos(theta)). How is that ? For 21

Ok so basically if T-mg=mv^/r then T=mg+mv^/r , but how do I find v^2/r in terms of g and a trig function? Also how do you find the period with large oscillations?

Homework Statement A simple pendulum experiment is constructed from a point mass m attached to a pivot by a massless rod of length L in a constant gravitational field. The rod is released from an angle θ0 < π/2 at rest and the period of motion is found to be T0. Ignore air resistance and...

Ok I got gt^4-4v^2+4L^2=0. How is that?