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vovangra [49]
2 years ago
12

Assuming that each of the following objects is a typical example of its class, rank them by increasing density.

Physics
1 answer:
inysia [295]2 years ago
8 0

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

<u>Molecular cloud-</u> They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between<u> 10 to 50 Kelvin. </u>Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

<u>Interstellar cloud-</u> They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around <u>just 100 Kelvin</u> while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

<u>1 Msun star-</u> These stars have temperature anywhere between <u>5300 and 6000 Kelvin</u>. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.  

<u>1 Msun protostar-</u> protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between <u>2000-3000</u> kelvin and are accompanied by dust usually.

<u>Intercloud gas- </u>These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around <u>8000 Kelvin</u>.

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Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache
Nina [5.8K]

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

       R = v₀² sin 2θ / g

How we know the maximum height

      v_{f}² =v_{oy}² - 2 g y

      v_{f}= 0

      v_{oy} = √ 2 g y

      v_{oy} = √ 2 9.8 / 15

      v_{oy} = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = v_{oy} / vo

    vo = v_{oy} / sin θ

    vo = 1.14 / sin 60

    vo = 1.32 m / s

We calculate the range with the first equation

     R = 1.32² sin(2 60) / 30

    R = 0.0503 m

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1 year ago
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45
Svet_ta [14]

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

5 0
2 years ago
Classify the following as alkali metals, alkaline earth metals, transition elements, or inner transitional elements: calcium, go
Digiron [165]
Alkali metals : sodium , potassium
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6 0
1 year ago
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If you are anchored in a fixed spot, and a set of six waves pass underneath you during a 60 second time interval, what is the wa
Lunna [17]

Answer:

Explanation:

Six waves passes by under 60second interval

Since it has 6 waves,

Then, it correspond to 3 wavelength

Then,

The period of each Each wavelength is 60 / 3

Then,  period = 20seconds

From the relationship between frequency and period

f = 1 / T

f = 1 / 20

f = 0.05 Hz

5 0
1 year ago
Suppose I have an infinite plane of charge surrounded by air. What is the maximum charge density that can be placed on the surfa
gregori [183]

Answer:

53.1\mu C/m^2

Explanation:

We are given that

Electric field,E=3\times 10^6V/m

We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.

We know that

E=\frac{\sigma}{2\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}

Using the formula

3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}

\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}

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\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2

1\mu C=10^{-6} C

3 0
2 years ago
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