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azamat
3 years ago
13

What type of galaxy is shown?

Physics
2 answers:
dexar [7]3 years ago
8 0
Elliptical, because the shape of the galaxy isn’t like the others. It is unique to its own and doesn’t have another to compare to
Katena32 [7]3 years ago
7 0

Answer:

Elliptical

Explanation:

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1 mole of air undergoes a Carnot cycle. The hot reservoir is at 800 oC and the cold reservoir is at 25 oC. The pressure ranges b
BabaBlast [244]

Answer:

The net work produced is 30.37 KJ

The efficiency of cycle is 72.3%

Explanation:

For the net work produced, we have the formula:

Work = (Th - Tc)(Sh - Sc)(M)

Where,

Th = higher temperature = 800° C + 273 = 1073 k

Tc = lower temperature = 25° C + 273 = 298 k

Sh = specific entropy at higher temperature

Sc = specific entropy at lower temperature

M = molar mass of air

Using, ideal gas table to find entropy. The table is attached.

therefore,

Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)

Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)

<u>Work = 30.37 KJ</u>

Now, for the efficiency (n), we have a formula:

n = 1 - Tc/Th

n = 1 - (298 k)/(1073 k)

<u>n = 0.723 = 72.3 %</u>

5 0
4 years ago
In a solution of soda, the gaseous carbon dioxide is the solvent, and the liquid cola is the solute.
satela [25.4K]
I think its true but im not sure
4 0
3 years ago
To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from
Natali5045456 [20]
1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.
3 0
3 years ago
Read 2 more answers
Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa
Alex

Answer:

B=1.1636*10^{-3}T

Explanation:

Given data

d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7}  )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T

3 0
3 years ago
The force exerted by the wind on the sails of a sailboat is fsail = 410 n north. the water exerts a force of fkeel = 200 n east.
tangare [24]
The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse.
(200)² + (410)² = F²
40,000 + 168,100 = F²
√(208,100) = F
456.18 N = F

F= ma
456.18 N = (300 kg)a
456.18 / 300kg = a
1.52 m/s² = a

The sailboat will be heading North East. To  find the angle of the boats trajectory use inverse tangent function.
tan(Ф) = opposite/adjacent
arctan(opposite/adjacent) = Ф

arctan(410/200) = 64° North East

7 0
3 years ago
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