Answer:197.504 N
Explanation:
Given
Two Charges with magnitude Q experience a force of 12.344 N
at distance r
and we know Electrostatic force is given
Now the magnitude of charge is 2Q and is at a distance of 
F'=16F
F'=197.504 N
<h2>
After 26.28 seconds projectile returns 26.28 seconds.</h2>
Explanation:
Initial velocity = 450 ft/s = 137.16 m/s
Angle, θ = 70°
Consider the vertical motion of projectile,
When the projectile return to the ground we have
Displacement, s = 0 m
Acceleration, a = -9.81 m/s²
Initial velocity, u = 137.16 x sin70 = 128.89 m/s
Substituting in s = ut + 0.5 at²
s = ut + 0.5 at²
0 = 128.89 x t + 0.5 x (-9.81) x t²
t² - 26.28 t = 0
t ( t- 26.28) = 0
t = 0 s or t = 26.28 s
After 26.28 seconds projectile returns 26.28 seconds.
magnitude of the net force = mass x acceleraton
= 22 x 2.3
=50.6 N
Answer:
The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.
Explanation:
Initial velocity = Vo= 25 m/s
Final velocity = V = x
Acceleration= a = 6 m/s^2
time= t = 4 seconds
Appy the equation:
V = Vo + at
Replacing:
V = 25 + 6(4) = 25 + 24 = 49 m/s
