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...
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...
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...
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.
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.
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?
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...
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...
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...
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
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.
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.
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.
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?
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.
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...