Recent content by Aphrael

The problem states: A loop of area 4cm^2 has its plane parallel to the field lines of the magnetic field, B = 0.6t, as shown in the figure. The loop is pulled in the opposite direction of the field with a constant velocity of v = 6m/s. I am confused as to solve this because I can't...

I have a few general questions which I need some clarification on. 1. What is really the relationship between potentials and test charges? I know how they relate to capacitance, but not really to each other. For example would a charge of 5mC be drawn to a potential of -7V or 7V? 2. I know...

I was using the Fx equation. That was my first one. However I wasn't using Fy. I'm not sure how to use it in this senario. F(e) I was using as electric force. I don't have a theta key so I just used 'q' as a dummy variable.

I was then using q=(F(e)r^2/k)^1/2. I knew that the length of my string was 1.2 meters, so I used tan(42.6)(1.2)=1.10 for half the radius between the two balls. However when I plug all these numbers in I get 1/95x10^-11 which is too small.

Alright it sounds like we're on the same tract so I'll post my equations. F(x)=F(e)-Tsin(q)=0 F(e)=Tsin(q)=mgsin(q)=(.012)(9.81)sin(42.6)=.0797 N (I just realized my calculator was in Radians, not degrees from my Calc class, but let's please continue because I want to make sure I understand...

Mine differs just slightly. There are two balls that form a triangle and the angle taken is from the horizontal splitting the angle formed by the two strings, but the angles equal. I noticed your first equation uses the opposite sign of what I used, but is that significant enough to change my...

Alright thanks.

Alright I tried to do what you told me. x-direction: F(tension)cos(q)-F(electric)= 0 y-direction: F(tension)sin(q)-mg= 0 Solving for F(tension) I can then use F(electric)=kq^2/r^2 to solve for q. Q=(F(electric)r^2/k)^1/2. However something I am doing is wrong because when I plug in...

Since the balls are equal in charge the force is directed away from each other. F(y) =mg =(.012)(9.81) and F(x) =mgcos(q) =(.012)(9.81)cos(42.6) Is this even close?

I know that the forces acting on the balls are gravity and string tension, but I do not think they are moving. There is a diagram with the problem that shows the two balls hanging, each with a 42.6 degree angle from vertical. Doesn't this mean that they are suspended on some sort of wire and not...

Homework Statement Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is q = 42.6, and the charges on the two balls are equal, what is...

Okay I'm more confused now. I know about about the logorithmic scale thing. But I also know that since dB=10 log(I2/I1) Going down 3dB is equivalent to decreasing by a factor of two since 10 log(2)=dB. I also know that every 10dB the intensity doubles. So I don't understand how 10dB increases...

One more question. Since I used KE=PE to find the v2 would the total energy still be KE+PE or just one of them?

Oops, sorry. Well I did think 2.5 kg. :) Thank you for your help.

2.5 kg is the mass of the block and the mass of the bullet is 5 g or .005kg.