Recent content by zooboodoo

if we do a maclaurin series expansion on sin(x^2) can't we use that to find the integral of sin(x^2)dx?

Homework Statement Integral(sinx(x)dx/(cos^2(x)+cos(x)-2) Homework Equations The Attempt at a Solution What I tried to do first was factor the denominator, so i got (cos(x)-1)(cos(x)+2) from there, I set up my partial fractions equation trying to solve B(cos(x)-1) + A(cos(x)+2) =...

wah it was supposed to be /3 wasn't it :-x so frustraating

\int{{1-x}{sqrt{2x-x^2}}dx

oh no the initial integral is (1-x)Squareroot(2x-x^2) all in the numerator

Homework Statement Find the integral of (1-x)((2x-x^2)^.5) dx Homework Equations The Attempt at a Solution I am kind of hazy on U substitutions but I thought that was the right way to go here: let u=2x-x^2 du=2-2x (1/2) Integral of (u)^.5 du < i wasn't sure if I went wrong...

if its increasing asymptotically towards 0 would that say that GPE is always negative, and relatively decreasing? Can anyone shed some light on the relationship of the equations F=GMm/r^2 and if / how it can be related to GPE=mgh in regards to the r and h values if they are infact related at all

And another part of the problem I had is why gravity is necessarily/how it was defined as a conservative force; that part I will start looking up soon.

The individual parts make sense, on their own. I was thinking that at a point, some distance, r/h gravitational potential energy would actually stop increasing and be decreasing, concepts of the spaceship being at a very large distance and having an incredibly high gravitational potential...

Sorry, for some more clarity what I thought was that: g is getting smaller faster than h is getting bigger in the equation for Gravitational potential energy (GPE=mgh). This is why the math / concept didn't make sense to m; like when you put height from GPE=mgh into the equation F=GMm/r^2, h...

This came from a practice MCAT test and I thought about it for a while and was still confused, I have the correct answer along with the explanation the textbook gave and was wondering if anyone had any light to shed on this. I didn't want to post this in the homework section because I'm not...

Homework Statement A charge of -1.9 × 10-9 C is at the origin and a charge of 9.4 × 10-9 C is on the x-axis at x = 3 m. At what two locations on the x-axis (xpositive, xnegative) is the potential zero? xpositive = m Xpositive = .5m v=kq/x + (k(5Q/3-x))to solve for this value, .5...

Sigh, power of 10 error. It was the right equation, thanks though.

Is the person traveling equivelocity relevant to the wind in the gondola? Imagine being in a convertible if that helps you picture it intuitively.

You're not given the final speed are you? I don't think you necessarily need to calculate it either.