Recent content by jenzao

So you use ohms law for each resistor and doesn't matter if its in parallel or in series with other resistors? (thats what's confusing me)

How do you find the current through a resistor that is in parallel with 2 other nonequal resistors? I totally forgot how to do this. as an example, if E = 10V R1= 5ohm R2 = 10ohm R3 = 15ohm So, I know that 1/Req = 1/R1 + 1/R2 + 1/R3 = .533ohm and also know Inet = 10V/.533ohm =...

sorry, pic pending approval --here it is uploaded to flickr... http://www.flickr.com/photos/31088870@N02/3560817988/

Homework Statement A circuit composed of one battery and four resistors is arranged as shown in the attached pic. E = 19 V R1 = 12 W R2 = 2 W R3 = 15 W R4 = 2 W Calculate the current I1 through R1? Homework Equations V=IR (obviously) The Attempt at a...

r is the distance from the charge where the potential is zero isn't it?

Homework Statement Is this question on electric potential solvable? I've tried everything. any ideas? A charge of -1.9 × 10-9 C is at the origin and a charge of 9.4 × 10-9 C is on the x-axis at x = 3 m. At what two locations on the x-axis (xpositive, xnegative) is the potential zero...

Homework Statement Consider three charges at the vertices of an equilateral triangle of side a = 7 cm. qA = +2 µC, qB = +2 µC, and qC = -1 µC. (see attached picture) Find the x- and y- components of the electric force acting on qC Homework Equations i am doing everything right...

Homework Statement Four charges are placed on the corners of a rectangle. What is the resultant force on the positive charge (a = 1.5 m, b = 0.5 m, q = 1.1 × 10-9C)? (see attached drawing) Homework Equations Im having problems summing the forces using superposition The Attempt at a...

Thought experiment about pressure (intro physics) ..you have a U-shaped tube? that is open to the atmos on both the left and right side you fill it with water so the water level is the same on both sides. then you put a rubber stopper in one end (..the right end) (also the remaining air...

Pascal's Principle: "A pressure change applied to a confined fluid is transmitted undimished to every point in the fluid and the walls of the container" there is a downward force on the surface of the water from the atmosphere. Therefore this change is transmitted everywhere undimished in...

OK is this correct then, (using my same example) the pressure on right (stoppered end) is same as pressure on left (open end) because the atm pressure from left side is transmitted equally throughout the liquid (pascals law), so there is upward force on right side of 101325Pa. But this force is...

no its never isolated because its a U-shaped tube. So are you saying that the atm pressure is transmitted up the other side of tube? thanks

Please tell me if I am understanding this correctly, or please point out where I am going wrong... imagine you have an open ended u-shaped tube filled with water. The pressure on left side of the U is the same as pressure at right side, because they are both open to atmospheric pressure...

this was explained really well, thank you. just quick follow-up.. Is gauge pressure simply pressure minus the air pressure? (101325Pa)?

Why is air pressure always factored in when calculating pressure at a given depth of water? Why is the force from air pushing down on the surface of water the same at 1m as it is at 100m? there is no air underwater, so why do we still need this value when doing pressure-depth calculations?