Recent content by Sink41

1. A parachutist of mass m falls freely until his parachute opens. When it is open he experiences an upward resistance kv where v is his speed and k is a positive constant. The parachutist falls from rest freely under gravity for a time \frac{m}{2k} and then opens his parachute. Prove that the...

If F = ma then the particle would speed off in one direction, unless it ended up or started at O with a velocity of zero. If F = -ma then you would get harmonic motion... I thought it might be plus or minus because in the question it didnt say whether the force was towards or away from O. I was...

Question copied word for word 1. A particle of mass m moves in a straight line under the action of force F where F = ms, s being the displacement of the particle from O, a fixed point on the line. When s = -a the velocity of the particle is u. Find the velocity of the particle when s = 0...

Ok, got an answer of 1.81 thanks for the help

Doesnt the normal force come out at a 90degre angle to whatever an object is resting on? Then R would change as you moved the ladder up the slope. I am confused whether you ment maximum force of friction as in when the ladder is at the bottom of the slope or whether you are saying that R is...

I already have both of those? torque: (1/2)(40g)(L)cos(theta2) = (L)(S)sin(theta2)horizontal: S + Rsin(theta1) = Fcos(theta1) (x axis)I just tried cancelling everything down again and got the two equations again: http://img20.imageshack.us/img20/8880/equationcy1.th.png andL = ((1 + (1 -...

Basically wondering what shortest possible length of this ladder is, one that reaches 2 meters high. The ground its on is quarter of a circle with radius 1 with an extra length of 1 on top. Weight is 40 (so 40*gravity for force) and µ (coefficient of friction) between ground and ladder is 0.8...

I used a trial version of derive 6 and plugged x^y=y^x straight in. It only plotted for positive values and went a bit crazy at high values of x and y. I think it was fairly accurate though. I remember putting one of the values for the non x=y line into my calculator and it working.

This wasnt set for homework just something i thought up so i hope this is the right place to post? Anyway when i put x^y = y^x (or also lnx/lny = x/y) in a graphing program i realized that there were two lines, y = x and another line that looked a bit like y = e^(e/x) but wasnt. So this...

Ok so how do i intergrate \int \cos(u) | \cos(u) | du ? btw i found a trial of a program called "Derive 6" on a demo disk i got from a maths course. It intergrated \int (1 - x^2)^\frac{1}{2} dx like this (click on thumbnail for...

instead of \int \cos^2 u du i should have \int \cos(u) | \cos(u) | du ?

OK... so when i change cosu to sinu... since i have (\cos^{2}(u))^{\frac{1}{2}}=(1 - \sin^{2}(u))^{\frac{1}{2}} then i get (|1 - sin{2}(u)|)^\frac{1}{2} out of it? so the answer is \frac {x} {2} (|1 - x^{2}|)^\frac {1} {2} + \frac {\sin^{-1}(x)} {2}

EDIT: tex is a work in progress... again :\ I'm guessing due to the fact -x and x give the same answer \sqrt{x^{2}}=|x| So (1-(X^2))^\frac {1}{2} = (1-(|x| ^2))^\frac {1}{2} and answer is \frac {x} {2} (1 - |x| ^2)^\frac {1} {2} + \frac {\cosec x} {2}...

Is that a subtle hint to stick the natural log function in there? because if you say u = 1 - x^2 then \int \frac{u^\frac{1}{2}}{(4 - 4u)^\frac{1}{2}} du which looks sort of f'(x)/f(x) ish... ok tbh i have no clue why you said absolute value except it means always...

EDIT: my tex is a little broken trying to fix So i want to intergrate \int (1 - x^2)^\frac{1}{2} dx i start off by saying \sin u = x so \frac {dx} {du} = \cos u then \int (1 - x^2)^\frac{1}{2} \cos u du which is \int \cos^2 u du...