Newton's Second Law Explained - Learn the Basics

don_anon25
Messages
35
Reaction score
0
Question answered! Thanks for the input!
 
Last edited:
Physics news on Phys.org
That resistive force is not the only force acting on the object (otherwise it'll just sit there). Don't forget gravity.
 
It moves along the x-axis...I forgot to mention that. So gravity is not taken into consideration?
 
If the object is initially at rest in a frame where the only force acting is proportional to velocity squared, then yes, it is obvious that the particle won't do anything. Somehow, I suspect that the question wan't meant to be that trivial.
 
I think that velocity to be v0, not zero. Otherwise, the problem is trivial. I just wanted to check and make sure I wasn't missing anything :smile:
 
Last edited:
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Replies
9
Views
1K
Replies
0
Views
2K
Replies
27
Views
2K
Replies
13
Views
951
Replies
37
Views
551
Replies
5
Views
2K
Replies
40
Views
4K
Back
Top