2. its the cross product of charge's velocity and the mf. If you point your fingers towards the direction of the velocity then point your palm in the direction of the mf (either up or down), then your thumb will be pointing to the direction of the force acted upon by the mf on the charge.
so...
oh i see. so the potential energy at C (105600 J) plus the kinetic energy at C add up to the beginning potential energy.
well then so is the answer
-110000J = 862.4x J
:confused: I am phys newbie so don't get mad at me but don't you have to calculate the work done by gravity between A and C?
W[g] = mgd = 220(9.8)(50-30) = 105600
so the potential energy at C as well as kinetic energy at D will be 105600.
I got -105600 for the change in KE between D and E. I got 122.45 as the distance beyond D when sled stops.
isn't work done by gravity between D and E zero? since the force vector is perpendicular to the displacement.
b.
positive acceleration equals speeding up
negative acceleration means slowing down and eventually opposite direction
0 acceleration is constant speed so nothing happens
current velocity doesn't matter.
There is a problem I am looking at and in its explanations for the answer it says
"...We know the sum of forces acting on m is T-mg which is equal to ma. Therefore, T=m(g-a)..."
um...Shouldn't T=m(g+a)?
you mean -DeltaU = -(Ufinal - U).
but what is the difference between a hand picking up an object against gravity, and a rocketbooster boosting a rocket against gravity?
shouldn't they be using the same formula?
but what I am seeing is the first problem uses W=-DeltaU and the latter uses...
Here is the question that I have no idea what is happening
"What is the change in potential energy of a particle of charge +q that is brougt from a distance of 3r to a distance of 2r by a particle of charge -q? "
Here's what I tried
-W = deltaU
W = qEd
W = (kq1q2/r^2) d
d= 3r - 2r = r...
W = U2-U1 = -G(m1m2)/2r - -Gm1m2/r
in the book it said potential grav. energy is -G(m1m2)/r where r is displacement and m1 is object mass, m2 is Earth mass
Hmm...im not sure
but here is the actual problem
A satellite of mass m is launched from the surface of the Earth into an orbit of radius 2r, where r is the radius of the Earth. How much work is done to get it into orbit?
Hi, I haven't taken physics yet, but I am reading a Sparknotes Physics book for fun, and there's something that i don't understand
In one part of the book, when it was talking about basic energy stuff it said that - ΔU = W...