How much force is needed to stop a 120-kg ice-hockey player if he decelerates at 20 m/s²?

1 answer:

3 0

ANY force at all, acting in the right direction, will stop him.
But smaller forces will take longer.

In order to either speed him up or slow him down at the rate
of 20 m/s every second, the force required is

(mass) x (acceleration) = (120 kg) x (20 m/s²)

= (2400 kg-m/s²) = 2400 N .

The force can either increase or decrease his speed at that rate,
depending on the direction in which it acts on him.

If the force is exerted opposite to the direction in which he's already
moving, then it can stop him if it continues long enough.

You might be interested in

Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put

Volgvan

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

8 0

2 years ago

Read 2 more answers

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$