If there is no gravity what will happen to earth?

PLEASE HELP ASAP WILL GIVE BRAINLIEST

stealth61 [152]

Explanation

(m) is measured in kilograms (kg)

<h2>(F) is measured in newtons (N)</h2>

<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>

4 0

3 years ago

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What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:

Sedbober [7]

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

$F_g = \frac{Gm_em_o}{r^2_e}$

Where,

G = Universal gravitational constant

$m_e$ = Mass of Earth

$r_e$ = Distance between object and center of earth

$m_o$ = Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

$F_g = m_o g$

Equation the two expression we have

$m_o g = \frac{Gm_em_o}{r_e^2}$

$g = \frac{Gm_e}{r_e^2}$

This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

$g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}$

$g = 9.8m/s^2$

The value of composite constant is $9.8m/s^2$ . Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

8 0

4 years ago

A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m

77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}$

$-3i+2j=3.0\times v_{2}$

$v_{2}=-i+0.66j$

The direction of the momentum

$tan\theta=\dfrac{0.66}{-1}$

$\theta=tan^{-1}\dfrac{0.66}{-1}$

$\theta=-33.42^{\circ}$

The direction of the momentum with respect to east

$\theta=180-33.42=146.58^{\circ}$

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0

3 years ago

Direction: Write TRUE if the statement is correct, and FALSE if it is not

Ostrovityanka [42]

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When light strikes a black surface, it is reflected by the surface and

nothing is absorbed This would be False because black absorbs the light not reflect

Mirrors are used in telescopes because they have the ability to reflect

True? Telescopes designed with mirrors avoid the problems of refracting telescopes. Because the light is reflected from the front surface only, flaws and bubbles within the glass do not affect the path of the light

Transmission in the passing of light through some materials

False? Some materials allow much of the light that falls on them to be transmitted through the material without being reflected. Materials that allow the transmission of light waves through them are called optically transparent. ... Materials which do not allow the transmission of any light wave frequencies are called opaque.

3 0

3 years ago

Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a dist

Elena L [17]

Answer:

Tangential speed of Mass revolution is greater

Explanation:

In the above question, we are given the following values:

Radius of Mars = 3.4 × 10^6 m

Mass of Mars = 6.4 × 10^23 kg

Mass of the sun = 2.0 × 10^30 kg at a distance of 2.3 × 10^11m.

Mar’s revolution around the sun = 687 days.

Time of Revolution = 24 hours and 37 minutes to complete one revolution.

Step 1

Find the tangential speed of rotation of Mars

Rotation of Mars = 2 × π × r

Radius of Mars = 3.4 × 10^6 m

Rotation of Mars = 2 × π × 3.4 × 10^6 m

Rotation of Mars = 21362830.044m

= 21.4 × 10^6 m

Tangential Speed of rotation for Mars = Rotation of Mars/ time

Time = 24 hrs 37 minutes

Converting to seconds = 88620s

Tangential Speed of rotation = 21.4 × 10^6 m/ 88620s

= 241.48047845m/s

= 241.5 m/s

Step 2

revolution of Mars = 2 × π × d

d = 2.3 × 10^11m.

= 2 × π × 2.3 × 10^11m = 1.45 × 10^12m

Revolution of Mars = 687 days

Converting 687 days to seconds = 59356800s

Tangential Speed of Revolution of Mars = Revolution of Mars/ times

Tangential Speed of Revolution = 1.45 × 10^12m/595356800s

Tangential Speed of Revolution = 24420m/s

Comparing the Tangential Speed of Rotation with the Tangential speed of Revolution, we can see the GREATER ONE IS THE TANGENTIAL SPEED OF REVOLUTION.

8 0

3 years ago

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