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Lorico [155]
3 years ago
14

Classify each planet as an inner planet as an inner planet or an putter planet

Physics
2 answers:
svet-max [94.6K]3 years ago
7 0

Answer:

Mars venus earth and Mercury are inner planets

Jupiter , Saturn , Uranus and Neptune are outer.

ololo11 [35]3 years ago
5 0

An inner planet (Or a terrestrial planet) is defined as a planet composed of rock and metals.

An outer planet (Or a Jovian planet) is defined as a gaseous planet with no stable surface.

With this being said, we can apply these definitions to all of the solar system's planets to identify which are inner and outer planets.

-Mercury: Mercury is composed of mostly metal. Because of this, it falls into the category of an inner planet.

-Venus: While Venus is known for having a very toxic gaseous atmosphere, the composition of the planet itself is made of rock and iron, classifying it as an inner planet.

-Earth: The planet we live on is made of rock & metal and as such is an inner planet.

-Mars: Mars, under its dusty exterior, is made of volcanic rock, becoming the last of the inner planet category.

-Jupiter: The largest planet in the solar system, Jupiter is mainly made of hydrogen and falls into the category of an outer planet.

-Saturn: Saturn is also composed of mainly hydrogen, making it an outer planet.

-Uranus: Liquid hydrogen and ammonia mainly make up Uranus, but methane is also in the atmosphere, which makes the planet appear blue. This is another outer planet.

-Neptune: The farthest away from the Sun, this outer planet is mainly composed of water, silicates, and various gases such as hydrogen and helium.

Therefore, the inner planets are Mercury, Venus, Earth, and Mars, and the outer planets are Jupiter, Saturn, Uranus, and Neptune.

<em>Hope this helped! :)</em>

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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is
Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
P = i^2R

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

P = (2.0^2)(50) = \boxed{\text{ B. }200 W}

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A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300
Illusion [34]
<h2>Answer:</h2>

\boxed{P=96.09MW}

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega

So we can calculate the power loss as:

P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}

Finally, the power loss due to resistance in the line is 96.09MW

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