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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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starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s
Anestetic [448]

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

a=25 m/s^2 is the acceleration

Substituting t = 29 s,

v=0+(25)(29)=725 m/s

Now we can find the distance covered in the last second of motion by using

s=ut+\frac{1}{2}at^2

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

a=25 m/s^2 is the acceleration

Substituting,

s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m

Note that the acceleration of 25 m/s^2 is not realistic for a car, but I still have used the data of the problem.

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6 0
3 years ago
A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7
Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

v_1 = \frac{m_2v_2}{m_1}

Apply the principle of conservation kinetic energy

K = \frac{1}{2}m_1v_1^2 +  \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\

v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14  \ m/s

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0
3 years ago
Question 11
notka56 [123]

The volume of the gas once it reaches the surface of water is 2 liters.

The volume of the air in balloon at depth of 100ft (30m) is 500ml.

The pressure at this point is 4 atm.

Assuming that the balloon have no compression by rubber of balloon the volume of air at the water surface is V.

The pressure at the surface of water is 1 atms.

As we know, from the ideal gas equation,

PV = nRT

Where,

P is the Pressure of gas,

V is the volume of the gas,

n is the number of moles,

R is the gas constant whose value is 0.082057 L atm mol-1 K-1,

T is the temperature.

Assuming that the temperature is constant,

We know,

PV = nRT

All quantities on the right side are constants,

So, we can write,

P₁V₁ = P₂V₂

Putting all the the values,

4(0.5) = 1V₂

V₂ = 2 Liters.

The volume of the air at the surface is 2 liters.

1 liter = 0.001 m

Hence,

2 liters = 0.002 m³

So the volume of air at the surface is 0.002m³.

To know more about Ideal gas Equation, visit,

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3 0
1 year ago
People who drink coffee are more likely to develop cancer
krok68 [10]

Answer:

The answer is False

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3 years ago
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When the net force of opposite forces is zero , the forces are
lorasvet [3.4K]
The answer is balanced
4 0
3 years ago
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