Answer: e. on the equator and in the Atlantic Ocean
Explanation:
Latitude and Longitude are geographical coordinates.
<u>Latitude</u> is the angular distance between the <u>equatorial line</u>, and a specific point on the Earth. It is measured in degrees and is represented according to the hemisphere in which the point is located, which can be north or south latitude.
In this sense latitude
refers to the equatorial line that divides the Earth in two hemispheres (North and South).
On the other hand, <u>Longitude</u> represents the specific <u>east–west </u>position of a point on the Earth's surface, being longitude
the prime meridian or Greenwich meridian.
So, if we have latitude
and longitude
(positive means it is to the East) we can already know the point is in the equator, and the option that bests describes the coordinates is e.
The topic here is momentum.
When a collision is said to be elastic, it means that the colliding objects now travel at their own new, indivual and distinct velocities, often in different directions.
So we write that as,
(mass of football player x velocity of football player) + (mass of referee x velocity of referee) = (mass of football player x velocity of football player) + (mass of referee x velocity of referee)
(M × 8) + (80 × 0) = (M× 0) + (80 × 5)
8M = 400
M = 50 kg
The correct answer should be:
Pascals Principal Is The Answer.
The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
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Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m