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Bezzdna [24]
3 years ago
7

Following are the different layers of the Sun's atmosphere. Rank them based on the order in which a probe would encounter them w

hen traveling from Earth to the Sun's surface, from first encountered to last?
A) Corona
B) Photosphere
C) Chromosphere
Physics
1 answer:
Ghella [55]3 years ago
3 0

Answer:

Sequence of layers to encounter while travelling from Earth to the surface of the sun are:

Option (A) - Corona

Option (C) - Chromosphere

Option (B) - Photosphere

Explanation:

  • Corona- It represents the extreme outer region of the surface of the sun. It is normally not visible because of the light emitted from the sun. So it is difficult to be seen from the naked eye, but it can be seen during the time of total solar eclipse. The temperature in this region is about several million degrees.
  • Chromosphere- It is a layer that lies between the Corona and the Photosphere. This layer has a thickness of about 2000 kilometers and the temperature in this layer ranges from about 6000°C to 20,000°C.
  • Photosphere- This layer is the bright visible layer of the sun which is comprised of plasma and dark and cool sunspots, that forms on the sun due to the emerging of the magnetic field of the sun from its surface.

Thus, the correct sequences are mentioned above.

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The crust is composed primarily of basalt and _____________.
german

Answer:

Granite

Explanation:

Trust me I learned this 2years ago

3 0
2 years ago
A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km
Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

v_f=v_o+at

Since the car starts at rest, you can write the following equation:

20=0+a(5) \\\\a=20\div 5=4 m/s^2

Now that you have the acceleration, you can do this:

d=v_o+\dfrac{1}{2}at^2

Once again, there is no initial velocity:

d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m

Hope this helps!

8 0
3 years ago
A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h
Varvara68 [4.7K]

Answer:

Explanation:

a = (vf - vi) / t

a = (50 - 90) / 10.0

a = -4 km/h/s(1000 m/km / 3600 s/h)

a = - 1.11 m/s²

5 0
2 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

3 0
3 years ago
C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-
mylen [45]

Answer:

C-14 has two more neutrons.

Explanation:

4 0
3 years ago
Read 2 more answers
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