The frequency of the radio station is

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength

is

where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
![[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a](https://tex.z-dn.net/?f=%5B%2890.0%29%280%29%2B%28.50%29%280%29%5D_b%3D%5B%2890.0%29%28v%29%2B%28.50%29%28-4.0%29%5D_a)
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
(I assume that the 4 directions north-south-east-west are meant with respect to the wire seen from the top.)
We can use the right-hand rule to understand the direction of the magnetic field generated by the wire. The thumb follows the direction of the current in the wire (upward), while the other fingers give the direction of the field in every point around the wire. Seen from the top, the field has an anti-clockwise direction. Therefore, if we take a point at east with respect to the wire, in this point the field has direction south.