Recent content by st3dent

thanks Thanks..seems to work now. I got it. Thanks for all your help.

I don't know what I'm doing wrong when taking this simple integral. The integral is: \int (1/x)(lnx)^2 dx \int (2/x)(lnx) dx 2 \int (lnx/x) dx Let u = lnx du/dx = 1/x dx = xdu 2 \int (u/x) xdu 2 \int (u) du 2 \int (u^2)/2 + C (2(lnx)^2)/2 + C (lnx)^2 +...

Gravity...but how do I solve this damn problem?

A model plane of mass .75 kg is flying at a constant speed in a horizontal circle connected to a 30 m cord and is at a height of 18 m. The other end of the cord is tethered to the ground as shown in the picture. The plane makes 4.4 revolutions per minute, and the force that the air exerts on...

The critical point on a graph is a point where: 1) The derivative of f is equal to zero. 2) The derivative of f does not exist.

Simple use the following to find the Average Value of a function: 1\b-a \int_a^b{f(x)dx}

Thanks Thanks for helping me.

Ok, i know the Force is 1312.67N and the velocity is 8.33 m/s the eqn for climbing the hill is mgsin4 + bv2 = Fperson i need to find the new v for going uphill. i know what mgsin4 and i know what b is...what if Fperson? I need to know Fperson to find the v going uphill. Thanks.

Male, Paris.

Ok...in response to the three equations..They make sense, but they assume that force is equal.. The answer I get is 8 m/s. The question asks if "the cyclist uses the same power, at what speed can he climb the hill" Does using the same power mean using the same force. I know that P = Work /...

Thank you.

Hello, I am difficulty solving this system of equations. Eqn1: 1.0416*10-21 = 4.75*10-27(x2) + 1.68*10-27(x1)(cos a) Eqn2: 2.827 = (x1)(sin a) Eqn3: 3.22896*10-16 = 8.4*10-28(x2)2 + 3.36*10-27(x1)2 I keep on getting equations with two variables in it. Can someone tell me how to...

I THINK I have a solution now. Since Ffrd = mg / 32 and F[sub]fr[sub] = \mu _kmgcos30 then \mu _kmgcos30 = mg/32 \mu _k = mg/32mgcos30 \mu _k = 1/32cos30 \mu _k = 0.036 Is that the right answer...THANKS!

I understand the following: Since the change in kinetic energy is zero, Ei = Ef mgh1 = mgh2 + Ffrd mgh1 - mgh2 = Ffrd mg(750 - 850) = Ffrd Since Ffrd = \mu _kFnd...

2. Two mountains are at elevations of 850m and 750m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one, with a total length of 3.2km and an average slope of 30 degrees, a) A skier starts from rest on the higher peak. At what speed will...