# Search results

1. ### Uncharged spherical conducting shell!

So +3q charge doesn't play any roll at all right?
2. ### Uncharged spherical conducting shell!

Yes, its E=q/[ε*area], where q is the charge enclosed.
3. ### Uncharged spherical conducting shell!

An uncharged spherical conducting shell surrounds charge -q at the center of the shell. Then a charge+ +3q is placed on the outside of the shell. When static equilibrium is reached, the charges on the inner and outer surfaces of the she are respecteively... +q,-q is the answer. Does the +3q...
4. ### Gauss Theory Question !

Yeah, its much different. Thanks!
5. ### Gauss Theory Question !

Im sorry, I meant 9 times more.
6. ### Gauss Theory Question !

So at the magnitude of the electric field is actually 4 times more for radius R than 2R?

E=kq/x^2
8. ### Gauss Theory Question !

Thanks Tanya for replying. But is the magnitude of E the same for both radius's?
9. ### Gauss Theory Question !

Two concentric imaginary spherical surface of radius R and 3R, respectively, surrounds a point charge -Q, located at the center of the surface. When compared to the electric flux I1 through the surface of radius R, the electric flux I2 through the surface 3R is. I know the answer is that...
10. ### Calculus 3 hw help

Yeah, me too. But the problem on the paper clearly states that. I'll bring it up to my professor. Thanks!
11. ### Calculus 3 hw help

let f(xy)=e^(2y)sin(pix), is f a solution to (fxy)^2 -fxx(fyy)=4pi^2 e^(4x) So I found fxx and fyy and fxy, which are -pi(e^(2y))sin(pix), 4sin(pix)e^(2y), 2pi(e^(2y))cos(pix) respectively, When i reduced everything i got e^4y(sin^2(pix))+pie^4ycos^2(pix)=pie^(4x). I am assuming it is...
12. ### Frequency Problem

I think I understand what you meant. In general, because the the source is moving with the medium, the air, it will have a higher frequency than when the source is static and receiver moving at the same velocity? Btw, thanks for being patient with me.
13. ### Frequency Problem

It is more than one. But I' still don't understand why they produce different numbers, unfortunately.
14. ### Frequency Problem

So fs/fd=va^2/(va^2-v^2)
15. ### Frequency Problem

Okay, so fs=fo(va)/(va-v) and fd=fo(va+v)/va and if fo and v are constants, lets say f=1 hz and v=2. I see now, so fs is larger. How can fs be larger though? Don't fs and fd have the same fo and v in the first place? This is a conceptual problem by the way.
16. ### Frequency Problem

Yes, sorry i meant higher frequency. So if I am correct, what you explained in the equations and situation shows that they both have same frequency because the ratios of fs and fd are the same right?
17. ### Frequency Problem

Sorry, I'm still having trouble grasping the concept. So,t he source is moving toward a stationary receiver (1) f=fo(v/v-vs), and the the receiver is moving toward a stationary source, (2) f'=f(v+vs/v). As they are moving closer, wouldn't the the observer hear a stronger frequency? So I solve...
18. ### Frequency Problem

Its f'=f(v(+/-)vD)/(v(+/-)vS). As car 1 moves toward car 2 it would be a plus sign on the numerator. And as car 2 moves toward car 1, the sign will decrease in the denominator. In the beginning, car 1 moves toward car 2 , then stops, and records the frequency, f. Now car 2 moves toward car 1...
19. ### Frequency Problem

Car 1 moves toward car 2 with speed v. An observer in car 2 measures the frequency of the sound emitted by car 1's horn to be f. Now, car 1 remains stationary while car 2 moves toward car 1 with speed v. The observer in car 2 now measures the frequency of car 1's horn to be f '. The relationship...
20. ### Find the Quadric Equation

Thanks Tanya Sharma for clarifying.
21. ### Find the Quadric Equation

1. Find the equation of the quadric equation that is the set of all points hose sum of their distances from the two points (2,0,0) and (-2,0,0) is 6. 2.I was wondering if rad[(x-2)^2+(y^2)+(z^2)]+rad[(x+2)^2+(y^2)+(z^2)=6 is the equation of the quadric equation? -vu2
22. ### Finding the Semimajor Axis

Simon Bridge and D H, Sorry for the typo and late posts, been busy with school, but what D H corrected is right. My professor told me to come back tomorrow to look over it. So I will have an answer by tomorrow night and will see what went wrong.
23. ### Finding the Semimajor Axis

Simon Bridge, I talked to my professor today, and he said I went ahead. That's why I couldn't figure it out, so I will try to attempt it again today. Thanks for your help.
24. ### Finding the Semimajor Axis

Simon Bridge, Thanks for helping. The equation I found was from a yahoo recipient, who had the same problem. I will ask my professor for assistance today.
25. ### Finding the Semimajor Axis

Simon Bridge, Yeah, for the part v^2/GM, instead of G as Nm^2/kg^2, I used 6.67m^3/kgs^2. So it will cancel the kg and s^2 leaving meters on the bottom like the part 2/r. Am I wrong, or is the online homework question is making a mistake?
26. ### Finding the Semimajor Axis

Simon Bridge, I first converted everything to meters, and my final answer came out to be 2.93x10^11 meters which equates to 293x10^6 km. I don't know what is wrong.
27. ### Finding the Semimajor Axis

A planet in another solar system orbits a star with a mass of 4.0 ×1030 kg. At one point in its orbit it is 250×106 km from the star and is moving at 35km/s. Take the universal gravitational constant to be 6.67 × 10−11 m2/s2 · kg and calculate the semimajor axis of the planet’s orbit. The...
28. ### Throwing a Ball

Simon Bridge, Yeah that's what I figured too. Thanks for the lesson.
29. ### Throwing a Ball

Simon Bridge, Sorry, I was suppose to get back to you sooner. That's an interesting question. I'm not sure how I would approach it, to be honest.
30. ### Throwing a Ball

Oh I see, thanks!