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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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An automobile tire is filled to a pressure of 240.0 kPa early in the morning when the temperature is 15.0C. After the car is dri
yuradex [85]

Answer:

<h2> 242.5kPa</h2>

Explanation:

According to one of the gas laws, \frac{P1}{T1} =\frac{P2}{T2}

Given P1 = 240.0kPa, T1, 15.0°C, T2 = 18.0°C, P2 = ?

Substituting this values into the equation, we have;

\frac{240}{15+273} =\frac{P2}{18+273}\\\frac{240}{288} =\frac{P2}{291}\\

Cross multiplying we have;

288P2 = 240*291

P2 = \frac{240*291}{288} \\P2 = 242.5kPa

The new pressure is 242.5kPa

5 0
2 years ago
a 2.0*10^3 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction b
BlackZzzverrR [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>KE = 1/2mv^2 = (1/2)(2000)(2^2) = 4000 J This must equal the net work acting on the car. W=Fd The net force is 1140-950= 190N. so, d=W/F = 4000/190 = 21.05 m</span>
4 0
2 years ago
The air in tires can support a car because gases __________.
____ [38]
Because the air inside the tires is kept at high pressure.

In fact, the force applied by the tires upwards to counter-balance the weight of the car (pushing downwards) is 
F=pA
where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.
8 0
2 years ago
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
2 years ago
If a force of 26 N is exerted on two balls, one with a mass of 0.52 kg and the other with a mass of 0.78 kg, the ball with the m
gogolik [260]
False is the correct answer
6 0
2 years ago
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