If the forces are equal, at a distance equidistant it is not possible to act a pair on the body since both torques cancel each other. Being of the same magnitude and in the opposite direction, the sum of the torques will be zero.
From the information in the question, the acceleration of the cart is 2.55 ms-2.
We can find the acceleration using the following information from the question;
u = 0.9 m/s
v = 6 m/s
a = 2.0 seconds
But;
a = v - u/t
v = final velocity
u = initial velocity
t = time
a = 6 - 0.9/2 = 2.55 ms-2
Force= ma
m = mass
a = acceleration
F = 6 kg × 2.55 ms-2
F = 15.3 N
Learn more about acceleration: brainly.com/question/12134554
A ) v = v o - a t
0 = 22 - a · t
a · t = 22
d = v o · t - a t²/2
0.04 = 22 t - 22 t / 2
0.04 = 11 t
t = 0.04 : 11 = 0.003636 s
a = 22 / t
a = 6050 m/s²
F = m · a = 0.09 kg · 6050 m/s²
F ( target→arrow) = - 544.5 N
b ) F ( arrow→target ) = 544.5 N
c ) If the speed was doubled: v = 44 m/s;
F = a m
a = 6050 m/s²
a · t = 44
t = 6050 : 0.04
t = 0.007272 s
d = 44 t - 44 t/2 = 22 t
d = 22 · 0.007272
d = 0.16 m = 16 cm
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