Answer:
The correct answer is D.
D:The surface of the coating allows light from the room to pass through but blocks the light from the screen.
Explanation:
Glare is produced on a computer screen when light from some external source reflects on the screen.
Anti-glare coating do not absorb light to reduce glare but they actually reduce glare by encouraging the light from the room to pass through the screen so that less light is reflected. Polarized lenses absorbs light to reduce glare, not anti-glare coating.
Complete Question:
Given
at a point. What is the force per unit area at this point acting normal to the surface with
? Are there any shear stresses acting on this surface?
Answer:
Force per unit area, 
There are shear stresses acting on the surface since 
Explanation:
![\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%2612%2613%5C%5C12%2611%2615%5C%5C13%2615%2620%5Cend%7Barray%7D%5Cright%5D)
equation of the normal,
![\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]](https://tex.z-dn.net/?f=%5Cb%20n%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%5C%5C0%5C%5C%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
Traction vector on 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]](https://tex.z-dn.net/?f=T_n%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%2612%2613%5C%5C12%2611%2615%5C%5C13%2615%2620%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%5C%5C0%5C%5C%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
![T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]](https://tex.z-dn.net/?f=T_n%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B23%7D%7B%5Csqrt%7B2%7D%20%7D%5C%5C0%5C%5C%5Cfrac%7B27%7D%7B%5Csqrt%7B33%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)

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


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

![\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](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%20%5B%5Cfrac%7B23%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_x%20%2B%20%5Cfrac%7B27%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_y%20%2B%20%5Cfrac%7B33%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_z%5D%20-%2028%28%20%281%2F%20%5Csqrt%7B2%7D%20%29%20%5Cb%20e_x%20%2B%20%281%2F%20%5Csqrt%7B2%7D%29%20%5Cb%20e_z%29%5C%5C%5C%5C%5Ctau%20%3D%20%20%5B%5Cfrac%7B23%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_x%20%2B%20%5Cfrac%7B27%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_y%20%2B%20%5Cfrac%7B33%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_z%5D%20-%20%5B%20%2828%2F%20%5Csqrt%7B2%7D%20%29%20%5Cb%20e_x%20%2B%20%2828%2F%20%5Csqrt%7B2%7D%29%20%5Cb%20e_z%5D%5C%5C%5C%5C%5Ctau%20%3D%20%20%5Cfrac%7B-5%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_x%20%2B%20%5Cfrac%7B27%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_y%20%2B%20%5Cfrac%7B5%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Cb%20e_z)

Since
, there are shear stresses acting on the surface.
Answer:
Explanation:
The stored elastic energy in the spring helps the box to rise to a height of .75 m
So stored elastic energy = potential energy attained
1/2 k d² = mgh , k is spring constant , d is compression in spring , m is mass of box , h is height attained .
Putting the values
.5 x 220 x d² = 4 x 9.8 x .75
d² = .2673
d = .517 m
51.7 cm .
Both F's are forces, so µ must be unitless/dimensionless.
F[friction] = µ F[normal]
Or, more explicitly in terms of the units:
(Newtons) = (constant) (Newtons)
Answer:
Explanation:
The method that I will use to locate the position of an object of ordered pair of value (8, 4) is known as the graphical method. The ordered pair of value (8, 4) is known as the coordinates. In graphical method, the coordinate will be located on the Cartesian plane having both x and y axis. The graph will be a 2-dimensional graph with y-axis being along the vertical and the x-axis along the horizontal.
The coordinates (8, 4) means that the value of x is +8 and the value of y is +4. To locate the point P(8, 4) on the graph, we will have to locate both points on the positive x and y axis respectively and then mark out their point of intersection. This point of intersection will be the position of the object defined as ordered pair of value (8,4).