Recent content by BATBLady

Homework Statement The drawing shows four point charges. The value of q is 1.96 µC, and the distance d is 0.93 m. Find the total potential at the location P. Assume that the potential of a point charge is zero at infinity. Image of problem is attached. Homework Equations V=kq/r The...

Homework Statement Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +2.9 cm and the other (q2 = -22 µC) is at x2 = +9.2 cm. (a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the...

=0.14V which is the 0.5(0.28)V, excluding a ^2

Ok, that's 0.28V/2.63=Vf

I'm sorry this is taking so long. I missed two lectures because I was sick, which is probably why I'm having this issue.

plugging into equation: 0.5(0.28)V^2=0.28V/2.36 ...V's cancel 0.14^2 (does not)= 0.28/2.36 0.0196 (does not)= 0.1186 That's where I'm at. I think I'm putting the equations into the wrong things...but I'm not sure. Comparing these give over 100% of the KE from the original.

KEb=0.14V^2 KEb+B= 1.315Vf^2 the answer for the problem should be (KEb+B/KEb)*100 the V's cannot cancel because they are not equal. If they were, it'd be over 100% of the KEb. I took the liberty of multiplying out the 0.5

So, if I'm doing this right 0.5*0.28V=0.28/2.63 V=0.76 plugged into the original KE, KE=0.081 Do I then plug that number in for KEb+B then use that number to compute the percentage of the original KE that the larger mass has?

Ok, so conservation of momentum would mean that 0.28V= before (2.35+0.28)Vf= after thus 0.28V should = (2.35+0.28)Vf

Am I assuming the KE is 0 or that it is equal to the KE of just the bullet? If KE=0 then u=1.1467. If not, u= sqrt(KEb+B/(.5*(.28+2.35)))

Inelastic Collision, only masses known need help Homework Statement A projectile (mass = 0.28 kg) is fired at and embeds itself in a target (mass = 2.35 kg). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does...

The way I've figured it, it'd be: 9.8cos16/8.7=a=10.82 m/s2 I put the answer in the system and it doesn't work out. Where am I going wrong?

I've been trying to do that. I know I'm trying to find acceleration, thus the equations I'm working with should be the ones I gave. However, I need the mass for both of those and I'm not given it (at least that's where my line of thought is).

Homework Statement A sports car is accelerating up a hill that rises 16.0° above the horizontal. The coefficient of static friction between the wheels and the road is µs = 0.87. It is the static frictional force that propels the car forward. (a) What is the magnitude of the maximum...

Nevermind about the silly program...it was my fingers missing numbers on my calc. The numbers that I gave in the reply were the right ones. Thank you so much though!