All categories

3 years ago

How did particles in the solar nebula eventually form earth

Physics

1 answer:

Sladkaya [172]3 years ago

5 0

Answer:

At the high temperatures of the inner solar nebula, the small proto-planets were too hot to hold the volatile gases that dominated the solar nebula. These proto-planets were Earth, Mars, Venus, and Mercury.

Explanation:

The materials that accreted into the early Earth were probably added piecemeal, without and particular order. The early earth was very hot from gravitational compression, impacts and radioactive decay; the earth was partially molted. The denser metallic liquids sank to the center of the Earth and less denser silicate liquids rose to the top. In this way the Earth differentiated very quickly into a metallic, mostly iron core and a rocky silicate mantle.

You might be interested in

A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the

erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$

5 0

3 years ago

A body is accelerated continuously. What is the form of the graph

Andrej [43]

Answer:

If a body is constantly accelerating the graph would be one-quadrant

4 0

3 years ago

Can someone define 'work' for me please? I looked it up but there are a lot of different answers.

vekshin1

Answer:

Work- Activity involving mental or physical effort done in order to achieve a purpose or result.

Explanation:

This was the first definition from the dictionary on google :)

7 0

3 years ago

Read 2 more answers

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be $\sqrt{10^3}}$ years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

$1year * \frac{360degrees}{31.62years} = 11.38 degrees$

or about 12 degrees.

6 0

3 years ago

A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl

raketka [301]

The wavelength of the note is $\lambda = 39.1 cm = 0.391 m$ . Since the speed of the wave is the speed of sound, $c=344 m/s$ , the frequency of the note is
$f= \frac{c}{\lambda}=879.8 Hz$

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where $\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m$ is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
$L= \frac{1}{2f} \sqrt{ \frac{T}{\mu} } =0.31 m$

8 0

3 years ago