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3 years ago

The gravitational force between two objects increases as

Physics

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer: Their masses increases or the distance between them decreases

Explanation:

According to Newton's law of Gravitation, the gravitational force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where $G$ is the Gravitational Constant

As we can see, if one body increases its mass or if both bodies increase their mass, the gravitational force will increase as well.

In addition, is the distance between the two bodies is decreased, the gravitational force will increase.

Therefore:

The gravitational force between two objects increases as their mass increases or the distance between them decreases.

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In what type of job might being a difficult person not be much of a liability ?

zhannawk [14.2K]

A possible answer would be a stocker. a person who sorts thing or delivery person who just drops things off. there are little to no jobs with no human interaction. some good options for jobs where you work by yourself are:
Embalmer.Accountant.Travel Photographer.Tree Planter.Freelance Writer.Truck Driver.Data Scientist.Taxidermist. I HOPE THIS HELPS!!!

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3 years ago

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

2 years ago

Read 2 more answers

Two forces of magnitude 8N and 4N act at right angle to each other. The angle between the resultant and the 8N force is?

Marta_Voda [28]

Answer:

Draw the vector triangle (head to tail)

Let 8 be adjacent and 4 the opposite side

tan theta = 4 / 8 = .5

theta = 26.6 deg

4 0

2 years ago

A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

The gravitational force between two objects increases as​

The gravitational force between two objects increases as