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...

Homework Statement 1) In an isometric exercise a person places a hand on a scale and pushes vertically downward, keeping the forearm horizontal. This is possible because the triceps muscle applies an upward force M perpendicular to the arm, as the drawing indicates. The forearm weighs 23.0 N...

=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...