Answer:
The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Explanation:
Given that,
Point charge at origin = 2 nC
Second charge = 5 nC
Distance at x axis = 8 m
We need to calculate the electric field at the point x = 2 m
Using formula of electric field
Put the value into the formula
The direction is toward positive x- axis.
Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s