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TEA [102]
3 years ago
7

The inner planets are relatively small bodies composed mostly of rock and metals. Why did they develop this way instead of becom

ing gas giants like the outer planets?
Physics
1 answer:
sleet_krkn [62]3 years ago
5 0

The temperature of the early solar system explains why the inner planets are rocky and the outer ones are gaseous. As the gases coalesced to form a protosun, the temperature in the solar system rose. In the inner solar system temperatures were as high as 2000 K, while in the outer solar system it was as cool as 50 K. In the inner solar system, only substances with very high melting points would have remained solid. All the rest would have vaoprized. So the inner solar system objects are made of iron, silicon, magnesium, sulfer, aluminum, calcium and nickel. Many of these were present in compounds with oxygen. There were relatively few elements of any other kind in a solid state to form the inner planets. The inner planets are much smaller than the outer planets and because of this have relatively low gravity and were not able to attract large amounts of gas to their atmospheres. In the outer regions of the solar system where it was cooler, other elements like water and methane did not vaporize and were able to form the giant planets. These planets were more massive than the inner planets and were able to attract large amounts of hydrogen and helium, which is why they are composed mainly of hydrogen and helium, the most abundant elements in the solar system, and in the universe

https://lco.global/spacebook/planets-and-how-they-formed/



hope it helps

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Answer:

And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second

8 0
2 years ago
An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha
hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

F_D =F_g

By definition we know that the Drag force is defined as

F_D= \frac{1}{2} C_d \rho A V^2

Where,

C_d = Drag coefficient

\rho =Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

F_g = (m_1 +m_2) g

Where,

m_1 =Mass of sphere

m_2 =Mass of unknown object

Equating the two equations we have to

(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2

Re-arrange for m_2,

m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1

Our values are given by,

C_d = 0.5

\rho = 1.22Kg/m^3

V = 66.7m/s

m_1 = 3Kg

d= 32.7*10^{-2}m

r = 16.35*10^{-2}m

Replacing in the equation we have,

m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3

m_2 = 8.62Kg

<em>Therefore the mass of unknown object is 8.62Kg</em>

4 0
3 years ago
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

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cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

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