Join my me et.gogle .etj-dovn-kds

If you move two objects father Paul how does the force of gravity between the two objects change

Yanka [14]

The force of gravity gets stronger.

4 0

3 years ago

Express force in terms of base units

MariettaO [177]

<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

Base SI unit of mass = kg (kilogram)

Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

Force = mass × acceleration.

Force = kg × m/s²

<u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

$\mathfrak{Lucazz}$

4 0

2 years ago

Which of the following is correct definition of electrical energy

tatyana61 [14]

Answer:

Electrical energy. Jump to navigation Jump to search. Electrical energy is energy derived from electric potential energy or kinetic energy. When used loosely, electrical energy refers to energy that has been converted from electric potential energy

Explanation:

6 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is: