Here are all the formulas I have learned in my pre-AP physics class so far, I just finished the 1st semester:
They are in no particular order.
F_c = \frac {mv^2}{r}
a_c = \frac {v^2}{r}
F_F = \mu * F_N
C = 2 * \pi * r
V_f = V_i + at
V_f^2 = V_i^2 + 2ad
d = V_i * t + \frac...
yea - that's what i thought, but its not somehow, i think the teacher is wrong
im still not sure what to do on the second one,
Here's what i know:
F_{F} = \mu * F_{N} or F_{F} = \mu * 9.8m
and P.E. = .5mv^2
and W = Fd or W = mad
those are the only formulas i know of that might help with...
How much work must be done to stop a 980 kg car traveling at 108 km/h?
what i did was first convert the km/h to m/s:
108 * 1000 / 3600 = 30
then plug the mass and velocity into the kinetic energy forumula K.E. = \frac{1}{2} mv^2
\frac{1}{2}980(30^2) = 441000
I have checked and rechecked...
the velocity at its maximum point would be 0 because it is about to start falling back down.
using the formula V_{f}^{2} = V_{i}^{2} + 2ad, you can get the distance it travels upwards.
V_{f} =0
V_{i} =3
a = -9.8
Solve for d.
ok, i have a problem and its answer, but I am not sure how to get the answer:
Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 81 m long. The coefficient of kinetic friction...
found my mistake, was supposed to use cos instead of tan for V_{x} now I've got everything right, thanks a lot to everybody for all the help
ill fix my post above too.
here's what I've gotten:
V_{x} \approx 99.83
v_{y} \approx 75.23
used V_{y} to get time:
V_{f} = V_{i} + at
0 = 75.23 + (-9.8)t
t = 7.68
double it to get the real time
t=15.35
now, to get the distance:
vt = d
99.83(15.35) \approx 1532.39
ok so far i got everything i need for until...