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ahh clever
2. Gauss Law and Flux

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...
3. Gauss Law and Flux

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...
4. Double Integral

I am sorry my brain is just fried, gotta go take the test now. Wish me luck :/
5. Double Integral

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...
6. Double Integral

I am not following. I cant even begin to integrate it, regardless of the limits.
7. Double Integral

pretty sure the input got messed up and i couldn't fix it. limits are -x to 0 for y and 0 to 1 for x
8. Double Integral

\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...
9. Trick Triple Integral

\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.
10. Favourite Quotes

"It's so damn hot... milk was a bad choice..." -Ron Burgundy
11. Volume of solid bounded by 2 surfaces

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
12. Volume of solid bounded by 2 surfaces

(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?
13. Volume of solid bounded by 2 surfaces

I am starting to get it more. I am attempting the problem but am having trouble figuring out how exactly you get the right limits of integration.
14. Volume of solid bounded by 2 surfaces

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 =...
15. Energy Jumping

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...
16. Kinetic energy of positron/electron collision

Ahhh I am forgetting the basics! Forgot about rest mass energy. Thank You!
17. Kinetic energy of positron/electron collision

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...
18. Kinetic energy of positron/electron collision

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...
19. Is there no force acting on the box?

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...
20. Spring Question

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!
21. Spring Question

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...
22. Kinetic Energy, Conservation of Energy, Potential Energy, etc

Would be easier if I could access the diagram.. the link requires an account on that site.
23. Help! force question

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