Recent content by COCoNuT

yes my aim name is CDesign1111 what about you?

i hate masteringphysics, i perfer doing homework and turning it in so teachers can tell you what your doing wrong. "Mastering Physics here too, though I'm using Physics with Modern Physics for Scientists and Engineers, Third Edition by Wolfson and Pasachoff." same here, what hw# are you...

yup it's correct

oops, my bad agian, it's actually 150m from C to B lol. sorry...

i had a typo while changing latex into normal text. it's 1500m from C to B. sorry about that. how would i find the overall velocity of the swimmer? do i need to use the final velocity formula(v(t) = V(0) + at)? if the horizontal velocity is 5km, wouldn't it help the swimmer to get to...

k changed latex into normal text, i guess latex is not working right now.

A swimmer wants to cross a river, from point A to point B. The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 1500 m, and the speed V_r of the current in the river is 5 km/h. Suppose that the swimmer makes an angle of 45 degrees(0.785 radians) with respect to the line...

lol, i just figured out what you tried to tell me, I am a bit slow. wow, that helped a lot. my teacher goes really fast, and doesn't explain the little details and the book is just weird. i already solved 6 problems on my own(14 more to go) and got them all right. thank you agian, i love this forum

yea! thank you, i get it.

sorry to keep on bothering you guys, but how did you get those values?

sorry, you guys are right... well anyway, with the correct results... \frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C time to sub in theta... \frac{\frac{\sqrt{(4x^2-1)}}{x^3}}{16} + (\frac{\frac{\sqrt{(4x^2-1)}}{x^3})^3}{48} + C is that right? i drew a triangle and...

i don't see how you got that, i did the problem agian and got the same answer

\frac{1}{16} [\int sec^2\theta d\theta + \frac{1}{16}\int sec^2\theta tan^2\theta d\theta] my work: found the anti-derv. \frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2 ok now this is the part I am confused on... i know i need to draw a triangle and do something with...

dont take my word for it(im not good at physics, lol), but this is how i think your suppose to do it: x=590 (east) y=-635 (south, cause it's diving and negative because it's south) mag is just a^2+b^2=c^2 so... 590^2 + (-635)^2 = c^2 and direction is... arctan(y/x) = arctan...

ok so the problem should look like \int \frac{x^3}{tan \theta} right now. i don't know what to do next. i can't use u-du. can i use parts on this? if you set 2x= sec \theta then x= \frac{sec \theta}{2} then would i have to take the derv. of that? dx= 1/2*ln(sec(\theta)+tan(\theta))...