During long jump athlete runs before taking the jump by doing so he

Physics

1 answer:

Alenkasestr [34]1 year ago

5 0

A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

<h3>What is inertia?</h3>

Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.

Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

Learn more about inertia here: brainly.com/question/1140505

#SPJ1

You might be interested in

Do Planets with more mass have more gravity than planets with less mass

STatiana [176]

Here you go it was too long to type

8 0

2 years ago

What is Vatiable Velocity?

Allushta [10]

Answer:

Explanation: When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity. Example: A rotating fan at a constant speed has variable velocity, because of continuous change in direction.

7 0

2 years ago

I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit

Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

$M=5.97\cdot 10^{24} kg$ (mass of the Earth)

$m=3.0 kg$ (mass of the box)

$r=R=6.37\cdot 10^6 m$ (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

$F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N$

2) $g=9.8 m/s^2$

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

$\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2$

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol $g$ .