Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
Electricity is always going to take the path of least resistance to ground. The rubber in your shoes is not a conductor of electricity, therefore you are not completing the circuit and you don't get shocked. Your bare feet, on the other hand ARE conductors of electricity, so when you hold the wire, you complete the circuit and become the path of least resistance to ground... ZAP!
Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
Thhhhhhhhhhhhhhhhhhhhhhhhheeeeeeeeeeeeeeee answer is 1.56
