car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastward towards car 2 at a velocity of 1.5 m/s. calculate the force of Car 3 and Car 2. dunno if I phrased this right so I included a screenshot too. pls help asap will give brainliest

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = -

Where;

= The final velocity

= The initial velocity

For the the 265 kg mass, we have;

= 0.1627 m/s

= 1.5 m/s

Which gives the impulse a s F×Δt =  265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.