Recent content by AxM=Fam

I started with the similar thinking when it came down to releasing the load. The only difference is in thinking was that once the load started releasing then this would put all the load on the release mechanism side. I understand the load sharing due to the tension on the chain side; but then...

I came across a diagram of a release mechanism, which shows the operator how to double a release load. I have taking a basic physics course (not advanced) and I can't seem to make any sense of this. I am just trying to understand this diagram; is it correct? Sorry for the drawing; hopeful that...

1. The figure shows an end view of two long, parallel wires perpendicular to the xy-plane, each carrying a current I=5.00A but in opposite directions. What is the magnitude and direction of the B-field at point P that is produced by the current in the two wires? 2. B=(μo/4pi) integral...

Yes. That is correct to say that there is no charge since the electric field is uniform. Area was the same all around so the electric flux canceled each other out in the i, j, z directions. Q=0. Thanks for your help.

1. A box with a side of length L=2.54 cm is positioned with one of its corners at the origin of a rectangular coordinate system. There is a uniform electric field E= 4 N/C(i) + 5 N/C(j) + 6 N/C(k). How much charge is enclosed by the box? 2. I believe I should be using Gauss's Law...

I believe all your formulas and calculations look correct but I would think about why it would be necessary to add both masses when looking for the acceleration. It is only the maximum acceleration of the top block that is needed.

I think I am getting a better understanding of the problem. I was missing the charge for the electron which was e=-1.602 x 10^-19. I would just have to figure the net force of the two point charges acting on the electron. From there would I just have to divide the net force by the given electron...

1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)? 2. K=8.988X10^9 N(m^2/c^2) Fc= K(qs...