The fictitious force is a_0 but I don't know how to draw it on a fbd. Am I supposed to split it into x and y coordinates (I'm using the slope as the x-axis...)
I understand the idea of the Lewis Structure and the octet rule. What I don't know is where to place the atoms. And unfortunately my book doesn't explain it. They just draw the lewis structur and expect us to understand how they did it.
For example: Draw the lewis structure of H_2CO...
We have a slope, with mass M, at x degrees. Mass m is on the slope. The slope moves right at constant acceleration a_0. There is no friction between M and m, and no friction between M and ground.
Determine the Normal Force between M and m...
Homework Statement
Prove sinx+cosx=x has a single solution in [0, pi/2]
The Attempt at a Solution
This question seems very strange to me. How can I prove there is a single solution for every number between 0 and pi/2?
I'm not sure what formula to use.
I'm learning infinitesimal...
Thanks. I think you're right. Vy =0
but I don't get one last thing:
I now have the formula: y = 1/2at^2.
what do I do with the other formula I had:
r(t) = \frac{qE}{2m}t^2 + v_0t+ h = 0
Why would the y-component be zero? The particle was moving all along, it didn't start from rest.
and i totally forgot. What formula do I use to compute the time.
there's
F = -mg
x = x_0 + (v_0 cos-alpha)t and
y = y_0 + (v_0 sin-alpha)t - 1/2gt^2
(i forgot all my high school math...)
I don't think I can say anything about the initial velocity v0.
the formula I used is
r(t) = \frac{qE}{2M}t^2 + v_0 t + r_0 where v_0 is initial speed at t=0, and r_0 is initial placement at t=0. It is given that r_0 = h and we know that r(t) = 0 at y=0... That's how I got...
thanks
this is what I got but I'm still stuck:
r(t) = \frac{qE}{2m}t^2 + v_0 t + h = 0
how do I solve for t? Am I supposed to use the quadratic equation?
Did I use the correct equation? Also can I use E in general, or do I need to split it between Ex, Ey, and Ez?
Thanks!
[PLAIN]http://img23.imageshack.us/img23/421/electromagneticquestion.png
A positively charged particle, which charge q and mass m, reached an area with an electric field and a magnetic field. Electric field E is at y>0 and its direction is -y. Magnetic field B is at y<0 and its direction is...
I know I take their mass into account. that's why i have (b+c+d)a
but my question is referring to the net force. Do I take into account the force from the smaller mass or since the friction cancels it out (which is why it is not moving) I only take into account the large mass.
When I am calculating the net force, do I also include the force from the small mass (with mass b):
bg\sin \alpha
or I don't since the friction cancels it out?
as in either:
dg - cg\sin \alpha = (b+c+d)a
or
dg - cg\sin \alpha - bg\sin \alpha= (b+c+d)a
so basically the only force acting on the system is the weight from mass d. so:
m_dg = (b+c+d)a
correct?
and if there was friction between the mass and the slope it would be:
m_dg - f_k = (b+c+d)a