Which order shows decreasing momentum?

Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the

Ugo [173]

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

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5 0

1 year ago

Can anyone heelp me plzz plzz

Natali5045456 [20]

I think the answer is c) 1
Hope it helps and have a good day

4 0

2 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

so

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

Choose the appropriate descriptor for the term cubic centimeter .

Arada [10]

Answer:volume

Explanation:

4 0

3 years ago

Read 2 more answers

Two sources of light of wavelength 720 nm are 10 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

julia-pushkina [17]

Answer:

The value is $y = 0.00732 \ m$

Explanation:

From the question we are told that

The wavelength of each source is $\lambda = 720 \ nm = 720 *10^{-9} \ m$

The distance from the pinhole $D = 10 \ m$

The diameter is $d = 1.2 mm = \frac{1.2}{1000} = 0.0012 \ m$

Generally from Rayleigh's criterion we have that the distance between the sources of light for their diffraction patterns is mathematically represented as '

$y = \frac{ 1.22 * \lambda * D}{d}$

=> $y = \frac{ 1.22 * 720 *10^{-9}* 10 }{ 0.0012}$

=> $y = 0.00732 \ m$

6 0

3 years ago