The equation you have there (xf=xi+vi t+0.5 ai t^2) is only true for constant acceleration. You should workout the instantaneous velocity by integrating the acceleration with respect to time. THen do the same thing to get the position.
As for (1) and (3) I'm not sure where your questions points at. As for your second question, a weightless balloon filled with water is just water being limited to a confined space so, asking how much the balloon weights in water is equivalent to asking how much does water weights in water. I...
What you say is usually ok, but not for this paticular problem. All the current will flow only trough the resistor at the top of the circuit. As for the bottom part, all the current will flow trough the resistance-free cable.
There is a little trick here. Actually the circuit is quite simple (in fact, as simple as a circuit can be).
Hint: If you have two paths in a circuit and one of them has o resistance, all charges will flow trough that path.
There are only 2 things you don't know here, the acceleration and the final speed. All other things you do know. As the problem says the body starts moving from rest, so there's v0. Aside from that, you now it travels a distance of .25 miles, which all you need to now, simply define x0=0 which...
You have a problem with the units of measure. Thinking only of units what you have is
\left[\frac{n^2\pi^2\hbar^2}{2mL^2}\right]=\frac{(eV)^2s^2}{kg\cdot m^2}=\frac{eV^2}{J}
which is obvioulsy not in eV. You should work with h bar in joules and, at the end of your calculation, go to eV.
That's true in the region between the infinite wall and the delta potential, however, for x>0 you can assume that there are no waves propagating from the right since there's no potential to the right of the delta function were they could be reflected.
There is quite a common trick for this kinds of things, you can always express the dot product of two angular momentums as
J_1\cdot\J_2=\frac{1}{2}((J_1+J_2)^2-J_1^2-J_2^2)
The operators on the right hand side are easy to handle.
Hope this helps.
You should use chain rule I think, so
\frac{\partial }{\partial x}=\frac{\partial u}{\partial x}\frac{\partial}{\partial u}
If I understood your question, this is what you are looking for.
What did you do to determine the constants in the case where its open on either side? I'm guessing you can follow a similar procedure here for the open side. That plus normalization should be enough.
It's correct. You are not assuming that every state is equally likely, you know it from the fact that all are multiplied by the same constant (in this case 1) in the non normalized wave function.
This has no problem with the fact that acceleration is the derivative of the velocity. As it turns out, the velocity of a free falling body is v(t)=g*t, then:
a=dv/dt=g
which meas the acceleration is contant and equal to g. On the other hand, if you evaluate the velocity a t=0, then...
Well, the electric field always points in the direction positive charges would move under its action.
So te two positive charges should move a positive charge away from them, and the two negative ones should move it towards them. Think of it a little, watch the drawing and its pretty clear that...
You can actually solve this square without using vectors, just some geometry.
You have four electric fields (one for each charge). Two of them point to the lower right corner and the other two point to the lower left corner. All of this fields can be easily computed using the formula you gave...
Hi.
You don't need the charge at the point P. As you well said, the formula for electric Field is E=kQ/R. Only one charge appears there and it is the charge that produces the field.
You are probably confused with Coulomb's law that says the force excerted by a charge Q on a charge q is...
As you computed cos(θ)=0.03/0.05=0.6. And thats it. So then
Fy=7.2 X 10e6 N cosθ=7.2 X 10e6 X 0.6 N
As you see you never get (because you don't want to) to know the exact value of the angle θ, but there's no need to, because you already know its cosine.
As you said, this time the sign would be possitive. However tou have overlookes a couple of things.
The triangle formed by the charges is not an equilateral one, so the angle is not 60º. Actually it doesn't matter the actual value of the angle, since you only want to know its cosine, but you...
I think you got this wrong.
In order to look for the sign of the charge, you should think about in what direction it should excert the force. Since the two bottom charges are postive, the "push" the central charge upwards, which means that, in order to get null net force, you need de upper...
It's kind off hard to answer this question whithout looking at a diagram, but, by convension, the current has always the direction of the movement of positive charges, which means that, when the capacitor is being charged, the current moves "away" of the positive pole on the battery.
Whether...
Maybe I didn't understand the problem, but... lets see...
When the window is closed, the normal to the surface points outside of the screen (to your face), and the magnetic field is pointing up, so the angle between those two is 90 degrees. As the window is opened this doesn't change, the angle...
I don't really now this, but probably one way that the conservation of energy may have afected the sport is in the stick. You'd want a stick that is as efficient as possible when transfering kinetic energy to the puck, i.e. not oosing as much energy as sound or heat.
As I said, I'm not sure of...
Hi again alicia, i think you are very confused. The time here, is good for nothing. It doesn't matter how long it takes you to go upstairs, whether you can do t in 3 minutes or 3 hours is irrelevant, the work done is the same. You must approach the problem in a different manner, let me try to...
As you said, you already did the first part of this problem, wich is to find the change in kinetic energy (This isn't just a random question, you're gonna need this for the second part).
The bullet leaves the block with less kinetic energy because of the work the block does. Since the bullet...
As you mention W=\vec{F}\cdot \vec{d}, time doesn't appear there, and that's because the work done by a force is independent of the time length of the force action.
The meassurement of time it takes you to go upstairs is not really that useful. Try to approach the problem in a different way...
Hi Peter, i think you are a bit confused.
As you say, we are interested in the flux enclosed, wich means we are interested in that part of the magnetic field that goes through the ring rather than paralel to it.
Think of it this way. The flux is max when the magnetic field is perpendicular to...