Recent content by A(s)

true, that makes sense. Thank you

No problem : )

lol flat hill... i meant flat ground

Ya, that is the answer I get as well. The trick to the problem was definitely remembering that Fnet = 0 at constant velocity. That would have probably given you the extra equation you would have needed to not have too many variables

The only thing that bothers me is that in the first hill, Fn = Mg + X there must be some x that counters gravity to make the Fn = 0 What is x? Fn = Mg would be a car traveling on a flat hill.

stupid typos heh

Also, i think i missed a negative sign [980 - F*sin(30)]*(.330) - F*Cos(30) = 0

Do you follow the reasoning?

The box moves at a constant velocity when the force of kinetic friction equals the force of horizontal pull (net horizontal force = 0). You know for this problem that your net normal force is 980 - F*sin(30). so your force of friction in the negative horizontal direction, plus your Force of...

A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the...

that helps a bunch, thanks

i suppose i am just wrong then... they seem really simmilar though

how are vectors irrelevant... with the doppler effect, you don't take absolute value and say that as a source of sound is approaching, the pitch decreases

ok, well even still, if the relative direction of the speed of light has no bearing, what if you pull out v^2 out of the square root. the negative velocity might apply. gamma = plus or minus c / [v(c^2 / v^2) -1)]if v is negative, you still get a negative value

to apply my previous statement directly to this problem... Y = b/ (b^2)^(1/2) does not equal 1 if b is negative... it equals -1