J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
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.
Their inferences are based on evidence that they collect during their investigations. Readers learn that scientists gather and interpret evidence and draw conclusions based on this evidence. ... Once scientists have gathered evidence, they use it to make inferences about the things they are investigating.
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
brainly.com/question/14548124
