Answer:
the answer is UV Radiation
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:
A
Explanation:
Hooke's law! F(spring)=-kx
There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.
Answer:
Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd
Explanation:
The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is
d = 12 + 9 = 21 yd
The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was
d = 15 - 3 = 12 yd
If the cross-section of a wire of fixed length is doubled, the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>