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vovangra [49]
1 year 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]1 year 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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Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0
1 year ago
In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which
Nesterboy [21]

Answer:

A

Explanation:

voltage between A and C is equal battery's voltage.

7 0
1 year ago
A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°
Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_{1} = 2.00\ L

Pressure P_{1}= 1.00\ atm

Pressure P_{2}= 70.0\ kPa

Temperature T_{1}= 20.0°C = 293\ K

Temperature T_{2}= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

\dfrac{PV}{T}=\ Constant

For both temperature,

\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}

Put the value into the formula

\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}

V_{2}=\dfrac{101.325\times2\times280}{293\times70}

V_{2}=2.766\ litre

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5 0
1 year ago
100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:
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Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0
2 years ago
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
1 year ago
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