A 1000kg car pushes a 1900kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4200N . Rolling friction can be neglected. a) What is the magnitude of the force of the car on the truck?b) What is the magnitude of the force of the truck on the car?

4200 N of force is acting on the truck by car and an equal amount of force is acting on car by truck.  

Explanation: This question is based on Newton's law of motion. It is related to the situation when the car just tends to move forward.

According to the Newton's third law of motion every action has an equal and opposite reaction.

Here car is exerting a pull (action) of 4200 N on the ground hence ground will also exert an equal amount of 4200 N force on the car (reaction).

The moment car just tends to move, the net acceleration acting on the car and the truck is zero.  

So, applying the Newton's second law of motion:

Forward force on the car i.e. F and T,tension force acting on the pulley used to connect car and truck  

F - T1 = ma

a = 0 (no net acceleration)

So, Tension force acting on the car by truck is T1, equals to F i.e. 4200 N.

Similarly applying the second law of motion on truck,

The only net force acting on the truck is T2 (tension force on truck by car). And T1 = T2 because both the tension forces are acting on a common pulley.

4200 N of force is acting on the truck by car and an equal amount of force is acting on car by truck.  

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

i think they are all correct

explanation:

they seem good