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masha68 [24]
3 years ago
14

What are two reasons why the terrestrial planets formed closer to the Sun after the supernova event that initiated the formation

of the solar system?
Physics
1 answer:
elixir [45]3 years ago
6 0
It is because of their gravitational pull .. Due to high gravitational force the planets are tend to move closer and they infact form closer to the sun. as the mass of the sun is very large.
and some of them which is also considered as outer planets like the titan the moon of the Saturn .. has to orbit the Saturn due to the gravitational force . its all because of the gravitational force between the planets and the sun .. which comprises to form solar system
You might be interested in
A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t
harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton v = 3.2 \times 10^{6}\frac{m}{s}

Mass of proton m = 1.67 \times 10^{-27} Kg

The force on the proton in magnetic field is given by,

  F = q (v \times B)

  F = qvB \sin \theta

But \sin 90 = 1    (∵ Force is perpendicular to the velocity so \theta = 90)

  F = qvB

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

  F = \frac{mv^{2} }{r}

Where r = radius but in our case 0.23 m, q = 1.6 \times 10^{-19} C

By comparing above two equation,

  B = \frac{mv}{qr}

  B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6}  }{1.6 \times 10^{-19} \times 0.23 }

  B = 0.145 T

5 0
3 years ago
PLEASE HELP!!!
Arisa [49]

Answer:

Each type of electromagnetic radiation have been matched to their descriptions as follows:

Ultraviolet light A) Given off by very hot objects, such as the sun

Gamma rays B) Given off by radioactive substances  

Microwaves C) Used in radar and to heat food

X‒rays D) Used in the medical field to see bone structure

Explanation:

PLS GIVE BRAINLEST

6 0
3 years ago
A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?
Alina [70]

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

7 0
3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
The MSDS for glacial acetic acid says that it is a flammable liquid that can severely burn any human tissue it comes in contact
Alex777 [14]

Answer: D) It should be handled in a fume hood, away from open flames.

3 0
3 years ago
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