I asked a friend after posting this. They said you use the area of the left side and multiply it by the electric field coming out of the slope as if it was just a rectangle and there was no slope at all. No need for any cosin or angles at all. I tried this and it gave me the right answer but I...
Homework Statement
The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box E1 = 90 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 120...
Would the limits still be 0 to 1 for x then -x to 0 for y? Im running on 3 hours of sleep so I feel completely lost sorry.
If so switching the order would only add an x to the equation then plug in a 1 for the x and a subtracting the 0 form of the equation giving me a single integral of just...
\int^{1}_{0}\int^{0}_{-x} \frac{ysin(pi*y^2)}{1+y} dydx
Not exactly sue how to start this. I know that I need to integrate with respect to y first then use that solution and integrate again with respect to x however I do not believe integrating the initial problem is possible. Is there another...
\int^{1}_{0}\int^{x/2}_{0}\frac{y}{(2y-1)\sqrt{1+y^2}}dydx
Most of my attempts at this problem fail pretty quickly. Not even my calculator knows what to do with this one.
Ah yes yes I get it now. However I believe it is supposed to be 2, not 4. The equations were Z squared not just Z. But I think I get it now. Thank you!
Edit: Also I am not sure why I said set z=0, not what I did, just a bit tired I guess
(x2 + y2)1/2 = z2
(x2 + y2)1/2 = 8 − z2
I substituted then set z=0 and solved for x and y to get x = +/- (y2-16)1/2 and y = +/- (x2-16)1/2
Are these my limits or did I go wrong somewhere along the line?
Just working on some practice problems. I missed a couple classes due to sickness and just need some extra help. If you could walk me through how to do these types of problems that would be amazing.
Homework Statement
Evaluate the volume of the solid bounded by the surfaces
(x2 + y2)1/2 =...
Homework Statement
Julie has a mass of 60 kg. In an experiment she crouched down, then jumped straight up. Her lab partners
measured the height of her center of mass above the floor at three instants: 1) 0.43 m when crouched
down; 2) 1.02 m just as her feet were leaving the floor; 3) the...
Ah that does make sense. However I am having a little trouble with the actual calculating of the photons energy. I figured with conservation of energy the initial kinetic energy would equal the final kinetic energy. I still do not fully understand how to calculate the final energy of the photons...
Homework Statement
A positron is a particle that has the same mass but opposite charge of an electron. An electron and a positron are shot directly toward each other by a particle accelerator. They start very far from each other, each with a kinetic energy of 5 × 10−14 J. When they collide...
Considering the force of gravity acting on the box in all situations, I would assume that answer A (no force acting on the box) is not an answer for any situation unless the box is floating off in space far away from any other objects.
However if the answer meant "In which situation (s) is...
I had a feeling I was forgetting something. We just learned about all that recently as well so I don't claim to be an expert ;) glad I could help though!
At the point where it is stretched fully all the energy is in the form of potential energy. At the equilibrium point all energy in Kinetic.
Let's start at the equilibrium. Energy is all Kinetic so..
E=(1/2)mv^2
E=(1/2)(0.256)0.746^2
E=.0712
Now since that energy is conserved the energy at the...
Since force friction is 0 and force normal cancels with the force going into the ramp, the force going down the ramp is your Fnet.
Fnet = mgsin(60)
ma = m(9.8)sin(60)
mass on each side cancels out.
a = (9.8)sin(60)
a = 8.49