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ddd [48]
3 years ago
12

The difference in the composition of the terrestrial planets and the gas giants can be explained by the different condensation

Physics
1 answer:
miv72 [106K]3 years ago
6 0
The inner planets are usually rocky because the gravitational pull is stronger closer to the star or in this case the sun. The dust and rocky particles that are left over after a super nova or in a nebula will tend to orbit closer to a proto-star when a solar system is in its early days. In our solar system these planets are Mercury, Venus, Earth and Mars. Gases are less dense and will be less affected by the pull of gravity because rocky particles have more mass. The outer planets are gas giants formed from clouds of gas that would be further out in the spinning disk around a proto-star. 
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Scientists can use spectral analysis of stars to determine which of the following?
RoseWind [281]
I do believe all of these but core elements can be determined by spectroscopy which includes the use of electromagnetic radiation. Both the surface and core temperature can be measured using light. Surface elements can be found because the absorption lines of different elements in the spectra of the star, but I haven't heard anything about using spectral analysis for core elements.
4 0
3 years ago
Read 2 more answers
(marking brainliest) pease help asap! both of the questions are in the pdf, and please let me know which is question one and whi
REY [17]
<h2><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em><em><u>S</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em></h2>

<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>

<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>

<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>

<em>Electric potential = 150 volts </em>

<em>so we know Power = </em>

<em>volt^2/Resistance = </em>

<em>=</em><em>(150^2/250)(ohms/volts)</em>

<em>=</em><em>(22500/250)watt = 9</em><em>0</em><em> </em><em>w</em><em>a</em><em>t</em><em>t</em><em> </em>

<em>4)Heat energy (H) = Power(P)×Time(t)</em>

<em>4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joul</em><em>e</em><em>s</em>

<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em>

4 0
3 years ago
On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin
kobusy [5.1K]

Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

      d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

      d = 1/5 = 0.2

For the network with 10 slits

      d = 1/10 = 0.1

let's calculate the separation (teat) for each one

      θ = sin⁻¹ (m λ / d)

for 5 slits

     θ₅ = sin⁻¹ (m λ 5)

for 10 slits

     θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

       I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0
3 years ago
Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa
bearhunter [10]

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

8 0
3 years ago
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