All categories

2 years ago

2. What gravitational attraction does the moon have on the earth? (mass of earth is 6 x

Physics

1 answer:

lions [1.4K]2 years ago

7 0

Answer:

Given that,

Mass of the Earth m

=6×10

Mass of the Moon m

=7.4×10

Distance between the Earth and the Moon d=3.84×10

km=3.84×10

Gravitational Constant G=6.7×10

−11

/kg

Now, by using Newton’s law of gravitation

(3.84×10

)

6.7×10

−11

×6×10

×7.4×10

14.8225×10

297.48×10

F=20.069×10

F=20.1×10

Hence, the gravitational force of attraction is 20.1×10

You might be interested in

Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 N

solong [7]

Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10 (g = 10 m/ $s^{2}$ )

m = $\frac{25}{10}$

= 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

5 0

2 years ago

I need help can someone help me pls

pychu [463]

Answer:

copper

Explanation:

4 0

3 years ago

Read 2 more answers

You have discovered a planet that is one-quarter the radius of Earth (Rp = 1/4R⊕) and one-half as massive (Mp = 1/2M⊕). How does

ExtremeBDS [4]

It's 8 times as much as before.

4 0

2 years ago

A non-_____ rock has interlocking grains with no specific pattern.

Kazeer [188]

A non <span>foliated </span>rock has interlocking grains with no specific pattern.

3 0

3 years ago

Read 2 more answers

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago