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icang [17]
3 years ago
14

When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o

f them is not obscured by interstellar dust and gas), their distribution was found to be
A) spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction.
B) in a relatively flat disk almost perpendicular to the plane of the Galaxy, with a relatively higher density of clusters toward its center.
C) concentrated in the plane of the Milky Way and clustered around the Sun's position, indicating that the Sun is close to the Galaxy's center.
D) uniformly distributed throughout space, with no concentration in any area of the Milky Way.
Physics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

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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.

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Answer:

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cupoosta [38]
I definitely agree with choise that you consider to be as a correct one. I am pretty sure that prganism is correct because this is a unit which ebrases all the mentioned points. And in simple words, the rest of options represent collective objects but you have to answer with sole one. Hope you find it helpful.
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