Because [math]\mathbf{F} = q\mathbf{E} + q\mathbf{v} \times \mathbf{B}[/math]
cross products are 0 when the vectors are parallel.
/sqt/ Stupid Questions Thread Superb Owl Edition
Nathaniel Kelly
Benjamin Lee
Who is the sexiest cartoon milf?
Hunter Sullivan
I'm having a lot of trouble with a capillary surfaces class. We're learning about differential geometry, and I'm being asked to calculate the mean curvature of this surface:
[math]x =
[/math]
How do I do it? The homework tells me I need to take the cross product of the derivatives of x with respect to alpha and beta, but tafter working out the cross product, I'm getting a really ugly vector that I'm not sure how to take the magnitude of.
Please help, sci. I have no idea what the fuck differential geometry even is.
Nathan Smith
Related question, how would you go about plotting a vector field / surface like this in Matlab?
Levi Harris
Actually I think it's just cause the overall change in height relative to point 1 is decreasing to the mercury, though I'm not even sure what is being used, the top of h3 I guess since that's the total transition?
Ayden Martin
hydraulic head = elevation head + pressure head
I think you've got pressure head covered so it's just where you're measuring from now
Logan Bell
...
Aaron Gonzalez
>Who is the sexiest CARTOON milf?
Did I say anime? No! Anime/Manga doesn't count. I've excluded it to Western Animation.
Aaron Miller
if its parallel I don't think its induces a current
thanks for the response user
Angel Fisher
Probability question. We have an access point that serves N clients. The probability that the access point sends a packet to one of the clients in any time slot is [math]p_a[/math]. Independently, the probability that any client attempts to send a packet to the access point in a time slot is [math]\frac{1}{N}[/math]. Packets are only successfully sent if only one packet is attempted to be sent over the network in any time slot (i.e. either the access point sends a packet to a client and no client tries to send to access point or only one client sends packet to access point and access point doesn't send).
I need to write two expressions: one for the average number of successful packet transmissions from the access point and one for the average number of successful packet transmissions to the access point in any time slot. So far I have:
[math]N_s = (1-\frac{1}{N})^Np_a[/math]
[math]N_r = \frac{1}{N}(1-\frac{1}{N})^{N-1}(1-p_a)[/math]
where [math]N_s[/math] is average successful packet transmissions sent from access point and [math]N_r[/math] is average successful packet transmissions received by access point. Do those look right?