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2 years ago

How does newtons law of the universal gravitation help scientist describe the universe

2 answers:

5 0

You need to know everything is made of atoms they are always colliding and they are never still because you'd think that every thing in space is still all the time even if it's not making visibly

Bumek [7]2 years ago

5 0

It starts to explain how the solar system stays "together" so to speak, and how the solar system fits into the Milky Way galaxy, and how the galaxies fit together to make a universe of galaxies.
It does this through the idea of the universal force of gravitational attraction (always attractive in N's case) saying that it depends on the masses of the objects concerned, their separation, and a universal gravity constant called G (different from g which is the acceleration of gravity about 10m/s^2)
F=Gm1m2/r^2 is the (inverse square law) force and it applies to all bodies in the universe.

You might be interested in

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

2 years ago

When an object is in circular motion it is constantly changing its velocity.

EleoNora [17]

Its tangential speed is constant although its velocity is changing. As the object changes direction, it results in a changing of positive and negative signs of the velocity. Although, the magnitude of the velocity (speed) is not changing.

8 0

2 years ago

If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity

agasfer [191]

Answer:

The answer is going to be C.

Explanation:

Trust me. Im an expert in physics

6 0

2 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

2 years ago

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

8090 [49]

Answer:

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass c.m= 1.6m

hence 1kg is 1.6m from the c.m

and x is 0.4m from the c.m

Taking moment about the c.m

clockwise moment equals anticlockwise moments

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0

2 years ago