Recent content by Hakins90

Oh dear... i posted too quickly. I just realized why it isn't. My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference. A = \pi (r+dr)^2 - \pi r^2 = \pi (r^2 + 2rdr + (dr)^2 - r^2) = \pi (2rdr + (dr)^2) I now...

Homework Statement Finding the area of a disc by integration of rings. Homework Equations A ring of radius r and thickness dr has an area of 2 \pi rdr. The Attempt at a Solution Why isn't it \pi (2rdr + (dr)^2)?

Hmmm I thought \frac {d}{dx} ( \frac 12 v^2) = 2 * \frac 12 * v = v and not v \frac {dv}{dx} Sorry if I am being stupid :D

The height dropped in the fall isn't 11m, it is (11+3)m. Does it work now?

Yeah you're right, mgh = Fd The distance that the force from the water acts through is 3m, so distance d = 3m. Stick that into the equation and you are done :D

Hmmm its really a maths question. In my textbook it says - " v \frac {dv}{dx} = \frac {d}{dx} ( \frac 12 v^2) by the chain rule." I can't see how they made this jump from the L.H.S. to the R.H.S. Thanks

Hmm I am not familiar with those equations... Work done is Force x Distance. Since all the work goes into the kinetic energy of the arrow - Work Done = Kinetic Energy Fd = 1/2*mv^2 Yeah?

First case - Fd = 1/2mv^2 v^2 = 2Fd/mv v = sqrt(2Fd/mv) 25 = sqrt(2Fd/mv) - Equation (1) Second case (2F)d = 1/2mv^2 v^2 = 4Fd/mv v = sqrt(4Fd/mv) = sqrt(2)*sqrt(2Fd/mv) = sqrt(2) * 25 (from (1)) v = 35.4 ms-1(3sf) I think :D