Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a = 
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²
is this a true of false cuz if so this it true
Newton's 2nd law of motion: Force = (mass) x (acceleration)
If you want to move a 7-kg object with an acceleration of 6 m/s²,
then you will need to push it with (7 x 6) = 42 newtons of force.
You don't mean the force of an object.
I think its maybe light energy
Draw a vector diagram. The net force on particle 1 = F12 + F13 + F14 These forces have to be added as vectors.
We will resolve our forces along the direction 1-4 F12 (tot) = -kQq / a^2 in the direction of particle 4 F12 = -kQq *sin (45) / a^2 F12 = -kQq /( a^2 * sqrt(2) )
By symetry this is the same as F13 F13 = -kQq /( a^2 * sqrt(2) )
F14 = -kQQ / (Sqrt(2)*a) ^ 2
For net force on particle 1 :
F12+F13+F14 = 0 -2kQq /( a^2 * sqrt(2) ) + -kQQ / (Sqrt(2)*a) ^ 2 = 0
Some simple manipulation should give you :
Q/q = -2 sqrt(2)
