The answer is A. Newton's third law of motion states that for every action, there is an equal and opposite reaction. A rocket exerts a large force on the gas that is in the rocket chamber (action). The gas thus exerts a large reaction force forward on the rocket (reaction). The large reaction force is called thrust.
Answer:
A. The forces are the same size and in opposite directions.
Explanation:
Just as an opposite number will cancel a number: -1 +1 = 0, so an opposite force will cancel a force, with the result that the net is zero.
Answer:
The speed is
and the direction is heading north.
Explanation:
In collisions the force exerted by the objects that collide is higher enough than the external forces that we can neglect that external forces, with that assumption we can use the conservation fo momentum law that states, final total momentum (pf) is equal initial total momentum (pi) if there’re not external forces or they are small enough to be neglected. Mathematically:

The total momentum is the sum of the momentum of each of the bodies we're dealing, in our case the moment of each car, then:

with pn the momentum of the 1000kg car heading north and ps the 800kg car heading south. Momentum is defined as mass times velocity, then:
(1)
It's important to note that when we talk about momentum and velocity direction matters, so we're are going to choose a system of reference where quantities pointing north are positive and pointing south are negative. So, the initial velocity of 1000 kg car is vni=5 m/s, initial velocity of 800 kg car is vsi=-4 m/s and the final velocity of 1000 kg car is vnf=-1 m/s. Now we can solve (1) for vsf and use the values we already have:

Because the sign is positive the direction is to heading north.
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m