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Answer:
Jace~
Plus, here is a anime image that might make your day happy. . .
The three ways a person can manipulate light
would be the following:,
filter, and the time the photograph is taken
<span>1.
</span>Angle
- <span>The </span>camera angle<span> <span>marks
the specific location at which the movie </span></span>camera<span> <span>or
video </span></span>camera<span> is
placed to take a shot.</span>
<span>2.
</span>Filter - Camera<span> <span>lens </span></span>filters<span> <span>still have many uses in digital photography,
and should be an important part of any photographer's </span></span>camera<span> bag.</span>
<span>3.
</span>Time
the photograph is taken - The golden hour, sometimes called the "magic
hour", is roughly the first hour of light after sunrise, and the last hour
of light before sunset, although the exact duration varies between seasons.
During these times the sun is low in the sky, producing a soft, diffused light
which is much more flattering than the harsh midday sun that so many of us are
used to shooting in.
I am hoping that these answers
have satisfied your queries and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.
<span>92.96 million mi..........</span>
Answer:
The curl is ![0 \hat x -z^2 \hat y -4xy \hat z](https://tex.z-dn.net/?f=0%20%5Chat%20x%20-z%5E2%20%5Chat%20y%20-4xy%20%5Chat%20z)
Explanation:
Given the vector function
![\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z](https://tex.z-dn.net/?f=%5Cvec%20A%20%28%5Cvec%20r%29%20%3D4x%5E3%20%5Chat%7Bx%7D-2x%5E2y%20%5Chat%20y%2Bxz%5E2%20%5Chat%20z)
We can calculate the curl using the definition
![\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|](https://tex.z-dn.net/?f=%5Cnabla%20%5Ctimes%20%5Cvec%20A%20%28%5Cvec%20r%20%29%20%3D%20%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%20x%26%5Chat%20y%26%5Chat%20z%5C%5C%5Cpartial%2F%5Cpartial%20x%26%5Cpartial%2F%5Cpartial%20y%26%5Cpartial%2F%5Cpartial%20z%5C%5CA_x%26X_y%26A_z%5Cend%7Barray%7D%5Cright%7C)
Thus for the exercise we will have
![\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|](https://tex.z-dn.net/?f=%5Cnabla%20%5Ctimes%20%5Cvec%20A%20%28%5Cvec%20r%20%29%20%3D%20%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%20x%26%5Chat%20y%26%5Chat%20z%5C%5C%5Cpartial%2F%5Cpartial%20x%26%5Cpartial%2F%5Cpartial%20y%26%5Cpartial%2F%5Cpartial%20z%5C%5C4x%5E3%26-2x%5E2y%26xz%5E2%5Cend%7Barray%7D%5Cright%7C)
So we will get
![\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z](https://tex.z-dn.net/?f=%5Cnabla%20%20%5Ctimes%20%5Cvec%20A%20%28%5Cvec%20r%20%29%3D%20%5Cleft%28%20%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%28xz%5E2%29-%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%28-2x%5E2y%29%5Cright%29%20%5Chat%20x%20-%20%5Cleft%28%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%28xz%5E2%29-%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%284x%5E3%29%20%5Cright%29%20%5Chat%20y%20%2B%20%5Cleft%28%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%28-2x%5E2y%29-%5Ccfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%284x%5E3%29%20%5Cright%29%20%5Chat%20z)
Working with the partial derivatives we get the curl
![\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z](https://tex.z-dn.net/?f=%5Cnabla%20%20%5Ctimes%20%5Cvec%20A%20%28%5Cvec%20r%20%29%3D0%20%5Chat%20x%20-z%5E2%20%5Chat%20y%20-4xy%20%5Chat%20z)
The distance travel is 69.5 meters.
<u>Explanation:</u>
Given datas are as follows
Speed = 27.8 meters / second
Time = 2.5 seconds
The formula to calculate the speed using distance and time is
Speed = Distance ÷ Time (units)
Then Distance = Speed × Time (units)
Distance = (27.8 × 2.5) meters
Distance = 69.50 meters
Therefore the distance travelled is 69.50 meters.