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murzikaleks [220]
2 years ago
10

How to find the density of air?

Physics
1 answer:
nikitadnepr [17]2 years ago
8 0
You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.
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A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o
bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = \omega=7.27\times 10^{-5} rad/s

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s

So, the velocity of the satellite is 3.08 km/s.

6 0
3 years ago
The area around a charged object that can exert a force on other charged objects is an electric ___
Ymorist [56]
Answer is:

Electric field.
8 0
3 years ago
What is the significance of a standard system of measurement?
dimulka [17.4K]

Because scientists all over the world are working together, looking for answers to the same questions, just as much as if they all worked in the same physical laboratory in the same building.  They need a way to share data and experimental results in a form that everyone can understand. ( D )

Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results.  That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

Now let's say that my results looked like this:

Result #1). 

The reaction stabilized when it reached the rate of 1.26 briligs per tove.

Result #2).

After running at that constant rate for 35 toves, a pile of product was produced whose mass was exactly 61.284 wibbles.

Result #3).

When the pile of product was allowed to settle for another 20 toves, it had spread out, and covered an area of 14.907 square filks.

Do YOU understand my results ?

All those other scientists would have a tough time trying to decide whether my results made sense.  And if they repeated my experiment, they would have no way to tell whether their results matched mine or not.

Without a standard system of measurement, and units that mean the same thing to everybody, us scientists literally could not communicate.


3 0
2 years ago
Read 2 more answers
Betelgeuse is how many times larger than the sun
Aleksandr-060686 [28]
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
3 0
3 years ago
(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass
11Alexandr11 [23.1K]

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants.  Mass of the Earth: 5.97 × 10^24  kg (assume a uniform mass distribution)  Radius of the Earth: 6371 km  Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?  

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

so  

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

6 0
3 years ago
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