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)
Read more about electric fields
brainly.com/question/15800304
#SPJ4
<span>Weight is directly proportional to mass.</span>
Answer
What makes up a perfect planet? It is the right distance from the Sun, it is protected from harmful solar radiation by its magnetic field, it is kept warm by an insulating atmosphere, and it has the right chemical ingredients for life, including water and carbon. Proportionate Ozone Layer and Light amount. According to the panspermia hypothesis, microscopic life—distributed by meteoroids, asteroids and other small Solar System bodies—may exist throughout the Universe. This is the perfect planet. In the end a perfect planet includes SUSTAINBLE DEVELOPEMNT IN EVERY ASPECT OF LIFE!
Explanation:
Answer:
Part a)

Part b)

Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction


now we will have force balance in Y direction


Part a)
so from above equations we have



Part b)
Now for tension in right string we will have


Power = work/time = (Force times distance)/time
= (30N *10.0m)/5.00s = 300/5 = 60 Watts