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3 years ago

On what bases are galaxies classified?

Physics

1 answer:

andreev551 [17]3 years ago

7 0

Based on the fact that the galaxy classification names all have to do with shape, your answer is C.

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Why does New and Full moons produce stronger tides.

serious [3.7K]

Because at the time of New and Full moons, the Earth, Sun,
and Moon are all lined up on the same line. So the effects
of the Sun and Moon combine to influence the tides in the
same direction.

4 0

4 years ago

If a chlorine atom gains or loses a valence electron, it becomes a charged

k0ka [10]

Answer: D. Chloride atom

6 0

4 years ago

A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai

Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air ( $\rho$ ), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$ , $M = 28.965\,\frac{kg}{kmol}$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$ , then the density of air is:

$\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}$

$\rho = 23.534\,\frac{kg}{m^{3}}$

The mass of air stored in the vessel is derived from definition of density. That is:

$m = \rho \cdot V$ (2)

Where $m$ is the mass, measured in kilograms.

If we know that $\rho = 23.534\,\frac{kg}{m^{3}}$ and $V = 10\,m^{3}$ , then the mass of air stored in the vessel is:

$m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})$

$m = 235.34\,kg$

The mass of air stored in the vessel is 235.34 kilograms.

7 0

3 years ago

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an

Zolol [24]

The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
We have then:
Horizontal = 9-9.2cos (58) = 4.124742769 N.
Vertical = 9.2sin (58) = 7.802042485 N
Then, the resulting net force is:
F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
Then by definition:
F = m * a
Clearing the acceleration:
a = F / m
a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
answer:
The magnitude of the body's acceleration is
2.941756275 m / s ^ 2

6 0

3 years ago

you arrive at an intersection with traffic lights that arent working because of a power outage .what do you do

ElenaW [278]

Use it as a four way stop.

6 0

3 years ago

Read 2 more answers