Forces acting on a ball as it is being squeezed

In which direction does a Newton’s third law reaction force act?

Norma-Jean [14]

<h2>Hello!</h2>

The answer is: B. In the opposite direction from the action force.

Newton's third law states that there is always a force acting in the opposite direction from the first applied force to an object, the second force will have the same magnitude but the opposite direction.

For example, when we sit on a chair, there is a force applied on the chair (our weight), at the same time, there is an equal force but with opposite direction going from the chair to us.

There are other two laws established by Newton, and they are related to the object's movement, acceleration, forces, masses, and velocity.

Have a nice day!

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3 years ago

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A parachutist's falls to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. Aft

Juliette [100K]

Answer:

$V_f=22.596\ m/sec$

Explanation:

<u>Physics Dynamics </u>

The second Newton's Law, states the acceleration of a body will depend on the net force applied to it and its mass. If an object is left on free air the net force acting on it is the gravitational force. It will continue to fall with accelerated motion until that force is changed.

The formulas needed to compute the physics dynamics magnitudes are

$F=ma$

$V_f=V_o+at$

$W=mg$

The variables involved are: F=net force, m=mass, a=acceleration, $V_f$ = final velocity, $V_o$ = initial velocity, t = time, W = Weigh, g= $9.8\ m/sec^2$

At first, only the gravitational force of 605 N acts on the parachutist. That net force is due to the parachutist's weigh. We can know the mass

$\displaystyle m=\frac{W}{g}=\frac{605N}{9.8m/sec^2}=61,735\ kg$

If we assume the initial speed is 0, then

$V_f=gt$

All variables are assumed to be positive downwards. So, when t=3 sec

$V_f=(9.8)(3)=29.4\ m/sec$

Right then an air resistance force of 665 N appears. The new net force is

$F=605N-665N=-60N$

The new acceleration will be

$\displaystyle m=\frac{F}{m}=\frac{-60N}{61,735\ kg}=-0.972\ m/sec^2$

The acceleration is now negative since it goes upward. We are required to compute the speed after 10 sec (not clear if it's after this last event or it comes from the initial condition). We assume those 10 sec come from the very beginning of the jump, so t=7 sec

$V_f=29.4\ m/s-0.972\ m/sec^2(7\ sec)=22.596\ m/sec$