Lenz's and Faraday's Law Induction. Help on resistor and switches!
Find the direction of the current in the resistor shown below a) at the instant the switch is closed b) after the switch has been closed for several minutes and c) at the instant the switch is opened...
A magnet is hung by a string and then placed near a wire as shown. When the switch is closed, the magnet rotates such that the ends of the magnet move as indicated by the arrows. At the instant the switch is closed determine:
a) the direction of the current through the wire...
Two long parallel conductors carry currents I1 = 3.00A and I2 = 3.00 A, both directed into the page as shown below. Determine the magnitude and direction of the resultant magnetic field at P.
B = uI/(2*3.14*r)
u = 1.26*10^-6 T*m/A
The Attempt at a...
The two wires shown below carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field a) at a point midway between the wires b) at point P1, 10.0 cm to the right of the wire on the right, and c) at...
From your statement, it seems that voltage and current are directly related. Increasing the current increases the voltage by using the equation, V = IR. However, there is also this other equation P = IV. This equation shows you that current and voltage are inversely related. How can this be so?
In this problem, box A and box B contain unknown combinations of light buls. Bulb 1 is identical to bulb 2. The batteries are ideal.
1. a) In the circuit A (see first attached pic) the voltage across bulb 1 and the voltage across box A are equal. What, if anything can...
It doesn't matter if you define force to be positive or negative. You can set the coordinate system whichever way you want. But if the direction of force and the displacement traveled are the same, then work is positive. In this case, force and distanced traveled are the same, I think. If I am...
Sorry, but I am just not getting this. E = k (q/r^2). As you can see, E depends on the distance between the charges acting on it. I know you're definitely right because the book says so, but it really did not explain much into it.
I thought electric field is always putting +1C of charge at the places on the points. Therefore, the closer it is to the positive side, the stronger the electric field, and the faster it would get repelled by the positive charge
Two very large sheets of charge are separated by a distance d. One sheet has a surface charge density +o and the other a surface charge density -o. A small region near the center of the sheets is shown.
1. Draw arrows on the diagram to indicate the direction of the electric...
A positive charge of magnitude qo is shown in the diagram below. Points B and C are a distance r away from the charge and point A is a distance 2r from it.
1. Would the absolute value of work done by an external agent in moving the same test charge from point B to C be...
Two charged rods, each with net charge -Qo, are held in place as shown in the diagram below
a) A small test charge -qo travels from point X toi point Y along the circular arc shown.
i. Draw an arrow at each point (X and Y) to show the direction of the electric force on...
Okay, so let's say we have a two bulb parallel circuit, each of them has 5? (whatever the units may be). If I plug that into the equation, it's going to be:
1/5 + 1/5 = 2.5
So putting the bulbs in parallel is going to decrease the resistence? Can anyone verify if I did the calculations...
Yeah, you're right. It seems like I have a problem of telling whether increasing bulbs in a parallel circuit would increase the resistence or not. I know that in a two bulb parallel, there is going to be more current going through the battery than the single bulb circuit. So there is going to be...
Difference between parallel and series circuit
Okay, I am really confused about parallel and series and their differences. I know how they are set up, but I do not know how their arrangements may change the circuit may result in a change in total resistence. From my lab, I know that connecting...
Gocha, but I still did not use the 20cm part from the question. I did the way you suggested:
For the x direction:
1*10^3N/C*q - T*sin 15 = 0
For the y direction:
Tcos15 - 0.002kg*9.8N/kg = 0
Solution: T = 0.0203N
q = 5.18*10^-6C
I still have doubt about my answer. The problem...
I put the tension in the x direction into the equation already. I do not think there is a need to worry about the y direction because the electric field is on the parallel direction and so net charge on the ball only exists on the x direction
A small 2.00-g plastic ball is suspended by a 20.0-cm-long string in a uniform electric field, as shown in the below diagram. If the ball is in equilibrium when the string makes a 15.0 degree angle with the vertical as indicated, what is the net charge on the ball? If you...
Okay, we are supposed to do a lab in class and compare it to the real results. However, for some reason, my lab result is very different from my theoretical value.
The lab is like this: I put a sheet of conductive paper with the dipole electrodes, one positive and one negative, of course...
When I did your method, I got this:
(P_i + kx/A)(V_i + Ax) = nRT_f
(1 + 2000x/0.01) + (5 + 0.01x) = 8.93
5 + 0.01x + 1000000x + 2000x^2 = 8.93
2000x^2 + 1000000 -3.9 = 0
x = (.1*10^6 +/- sqrt (1*10^12 + 31200)) / 4000
x = -500m, 3.9*10-6m
For this equation, I also got too small of a...
Hey denverdoc, I used your method and I got 1.79atm for the pressure and 8.95*10^-6m for x. I think the pressure is right but x is wrong, since the value is too little to be noticeable in a real life experiment. This is what I did:
N = S
PA = kx
Just by looking at the question, I know we have...
Hey, thanks pal. This really helped a lot. denverdoc, I completely understand your explanation. But Andrew, regarding your equation, I do not get how you derived this equation Pf = Pi + kx. If you can explain it to me, that would be great. Unit of pressure is in atm or N/m^2 and unit of kx is N...
A cylinder has its lid connected to a spring with k = 2000 N/m. The cylinder is filled with 5.00 L of gas witht he spring relaxed at a pressure of 1.00 atm and a temperature of 20.0C. (a) If the lid has a cross-sectional area of 0.0100 m2 and negligible mass, how high will...
I was having a lecture of buoyancy and I am confused about the meaning of displaced water. I asked my classmates about it and they said it is essentially the volume of the object when you put the object into the water. So now the question comes.
There are three objects...
oh, I see now, it took me a while to think about it until you told me to draw boxes. So just to verify if my thinking is correct. Although object A has a greater depth difference, the pressure exerted on the vertical surface is less than that of object B because it has a smaller surface area. On...
Two objects of the same mass and volume but different shape are suspended from strings in a tank of water as shown. Is there more tension force on the string by object A or B? Link to the picture can be found at http://students.washington.edu/cy1126/Buoyancy.JPG [Broken]...
A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path,(a) at the beginning of...
b)ohhh, I see what u mean by using the conservation of energy. So, it is mgh = 0.5mv^2 + mgh,
9.8 * 4 = 0.5 * v^2 + 9.8 * 1.5
v = 7m/s.
Plugging in the equation F = ma,
F = m * v^2 / r
F = 70kg * 7^2 / 4
F = 857N
Well the angle at 1.5m from the bottom of the circle would be arccos(2.5/4) = 0.89 rad --> pi/2 - 089 rad = 0.68 rad is the angle at that moment. Using the equation F = ma, you get F = m*V^2/r. However, I don't see how you can get velocity if you don't have time... Can someone help it out? I'm...
okay, I think I kinda know what ur saying. I'll give it an another shot
Ft = mvf
400N * 1.0s * 0.5 = 61.2 kg * vf
vf = 3.26 m/s
x = vit + 0.5at^2
x = 3.26 * 1s + 0.5*9.8*1^2
= 8.16 m
Is what I am doing correct? I don't get what u said about "You might want to use the equation that...
Alright, I think I'm getting this. So:
w = mg
600N/9.8N/kg = 61.2Kg
Ft = mvf - mvi
Ft = -mvi
(1.5s * 400N * .5) * 1.5s = -61.2kg * Vi (in this part, do you have to divide the force by two like I did since the area under the curve is roughly a right triangle?)
Vi = -7.35 m/s