All categories

3 years ago

which of the following refers to the motion of an object falling straight down with only the force of gravity acting on it

Physics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

Since you haven't provided any choices, then the answer is "Free Fall Motion."

Explanation:

In order to learn more about the answer, let's discuss what free fall motion is.

Free Fall- In Physics, this refers to any body motion that is acted upon solely by <u>"gravity."</u> The acceleration in free fall is always downward and there's the absence of other forces. Take note that the<em> acceleration should be the same and is independent of the object's mass. </em>This acceleration is called "acceleration due to gravity."

Gravity- This refers to the force that pulls any object towards the center of the earth.

<u>Examples of Objects in Free Fall Motion</u>

1. A ball dropped at the top of a building.

2. Dropping a coin from a table.

The ball and the coin are both in free fall motion because they are being pulled by gravity towards the earth. Their acceleration is also constant and there are no other forces acting upon them.

You might be interested in

1. What function does the skull serve for the skeleton?

Free_Kalibri [48]

1.c
2.b
3.a
4.c
5.b
I hop this helps

8 0

3 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

2 years ago

An object with a mass of 5.0 kg accelerates 2.8 m/s2 towards right when an unknown force is applied to it. What is the amount of

RoseWind [281]

Answer:

Explanation:

5 0

3 years ago

Any tips on how to get a good grasp of physics and chemistry?

Helga [31]

Answer: Take notes and research into the subject more get extra help like tutoring if needed.

Explanation: ...

4 0

2 years ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0

3 years ago

which of the following refers to the motion of an object falling straight down with only the force of gravity acting on it​

which of the following refers to the motion of an object falling straight down with only the force of gravity acting on it