Recent content by mr_sparxx

Thank you: that perfectly explains one of my doubts. I think that unfortunately it is the first case. This problem is for 17-18 years old who have just been introduced to the Bernouilli's principle. Although I have very basic notions of fluid mechanics I understand how the Benouilli's principle...

I am using some solved excercises to put homework to my students but I cannot understand the proposed solution for c) a) $$ Q= 400 l/min = 6.67 ·10^3 m^3/s $$ applying continuity equation: $$ Q = S_1 v_1 \Rightarrow v_1 = \frac Q S_1 = 4.19 m/s $$ This is exactly the same as the proposed...

I guess you mean ( mass is missing) ## \mathrm{d} A = \frac{1}{2}| \vec r \times \mathrm{d}\vec r| = \frac{1}{2}| \vec r \times \vec v|\mathrm{d}t =\frac{1}{2m}| \vec L |\mathrm{d}t ## Nice way of explaining it. Just what I was looking for. Thank you.

Many texts state that in an elliptic orbit you can find angular momentum magnitude as $$ L = r m v = m r^2 \frac {d \theta} {dt} $$ I wonder if $$ v = r \frac {d \theta} {dt} $$ is valid at every point. I understand this approximation in a circumference or radius r but what about an arc...

Actually what was wrong was my handwritten copy of the answer choices! :H I am so embarrased! :eek::sorry: Choices were: a) ##\frac {K q^2 sin 60º}{l^2}## b)##\frac {K q^2 cos 60º}{l^2}## c)##\frac {2 K q^2 sin 30º}{l^2}## d)##\frac {2 K q^2 cos 30º}{l^2}## And it is obviously c)... Sorry for...

Thank you for your answers and thoughts. I posted it here because I thought it was the most relevant place but, in fact, this problem comes from a past access exam paper one of my students brought to me. I was unable to give him an answer. Tomorrow I'll take a picture of it to be sure ...

The exact question is the one stated in part 1. I have to write a, b, c or d in the answer sheet and my result I am getting is different to all four choices: that's why I am so confused.

Thanks for your answer. I am sorry, I should have said it is a multiple choice question: a, b, c, or d are the answer options. I'll update the post. Could it be possible that the question is wrongly formulated or the answers wrong? This is the exact text of the question...

Homework Statement Three particles with the same charge q are hold together by three strings of the same length l forming an equilateral triangle. The tension in each string is: a) ## \frac {K q^2 cos 60º}{l^2}## b)## \frac {K q^2 cos 30º}{l^2}## c)## \frac {K q^2 sin 60º}{l^2}## or d)##...

So ## i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{abd} L_b L_d = i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{acb} L_c L_b = 0## So simple now... b is a dummy index! It is still difficult for me to realize what is "fixed" and what is a dummy index without writing down the summation...

Homework Statement Prove that ## [L_a,L_b L_b] =0 ## using Einstein summation convention.Homework Equations [/B] ## (1) [L_a,L_b] = i \hbar \epsilon_{abc} L_c ## ## (2) \epsilon_{abc} \epsilon_{auv} = \delta_{bu} \delta_{cv}- \delta_{bv} \delta_{cu}## ## (3) \epsilon_{abc} = \epsilon_{bca}...

I can't get this one correct either... I'll post a new thread anyway. Thanks a lot!

Seems ok to me. Just to complete the problem, try to express ## Q' ## in terms of given data (##a ## and ##Q##). I think 'unitary' is a bad translation... It is actually 'unit' vector: I'm sorry. I'm glad to help :)

Correct, since all contributions are parallel. Anyway, it does no harm. ; )

Yes but your integration limits are incorrect. You should be integrating (summing) contributions to the electric field of charges within the rod, not outside (##[a,3a]## is outside). Furthermore, you should have the distance from the charge ##dQ## to the point P (that is, ##r##) expressed as a...