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den301095 [7]
3 years ago
7

Nitrogen gas at standard atmospheric pressure 101.3 kPa, has a volume of 0.080 m3. If there are 3.6 mol of the gas, what is the

temperature?
Physics
1 answer:
marysya [2.9K]3 years ago
6 0

Answer:

271 K

Explanation:

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Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of
PolarNik [594]

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

6 0
2 years ago
Distinguish between friction force and motion?
Ksju [112]

Friction force is when you rub 2 things together and they get warm. Motion, on the other hand, is if your walking along the sidewalk - you hardly get warmer -------


Unless it's a colder day outside and you're walking SO you decide to rub your hands together to get warm, but if you were just walking , its motion and only motion - no friction :):)

6 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
3 years ago
determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the
Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0
3 years ago
28. Calculate the weight of a 1,789 kg car on the Earth.​
skad [1K]

Answer:

From the equation ΣF = ma, one can say that the mass is 1800 kg, and the acceleration of gravity is -9.8 meters. Multiply those together and it is -17640 newtons.

Explanation:

plz send a brainleast

3 0
3 years ago
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