Recent content by LR5

Summary:: Solving series-parallel circuits using current and voltage divider rule For the attached circuit, I need to determine the voltage drop across R4 using the voltage divider rule. So far, I have determined total resistance to be R = 383Ω, total current I = 0.03A, and I3 = 0.023A. I...

Thanks again. I don't know why this one had me so puzzled, though.

I sort of thought I had sine & cos mixed up.. Vpw = (-4.5sin30)x +(4.5cos30)y Vfp = 0x + (-1.5)y Vfw = -2.25x + 2.4y = 3.28m/s and 43.17 + 30 = 73.17 west of north

Lol, even the diagram? Please help...

Vpw = Vpf + Vfw Vfw = Vpw - Vpf Vfw = Vpw + Vfp Vpw = (-4.5cos30)x + (4.5sin30)y Vfp = 0x + (-1.5sin30)y Vfw = (-3.90)x + (2.25 - 0.75)y = 4.18m/s at (10.9 +30) = 40.9 degrees west of north

Thank you. I think I have it now.

I think I'm focusing on the person walking North relative to the ferry and I keep wanting to place that vector pointing north when it is in fact not north relative to the water.

?

Ok,that makes more sense. Thing is, when I change the formula from Vpw = Vpf + Vfw to Vfw = Vpw - Vpf, I'm not sure if I'm still supposed to switch the pw to wp (and therefore the direction) and I'm pretty sure the x component for Vpw would still be negative as well. So,lots of negatives...

This is what I have: Vfw = Vpf + Vpw = -Vfp + Vpw Vpf = 0x = (1.5)y Vpw = (-4.5 cos 30)x + (4.5 sin 30)y Vfw = -3.90x + 0.75y = 15.8 m/s at 10.9 degrees west of north

I drew a diagram which is basically the person walking north (positive x axis) from the origin and the then for the person relative to the water a vector 30 degrees from the north axis (to the left of the y axis) also starting at the origin. I am confused as to using the displacement diagrams...

(pw is person to water, pf is person to ferry, fw is ferry to water) I get Vpw = Vpf + Vfw. Therefore Vfw = Vpw + Vpf, which is Vfw = -Vwp + Vpf. I also have that the x component of Vpw is negative. I think I'm getting confused with either the order of the formula or the negative negatives (or...

Lol, ok. That's only half the time. Thanks, I have been making it a lot more complicated than it is.

So, I could just take the velocity of Ball 1 as the average velocity for Ball 2? (Vy = 13.5 sin 40)?

Vsquared=V(initial)squared + 2ax?