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NISA [10]
2 years ago
9

In which type of shoot is continuous lighting used?

Engineering
1 answer:
Fiesta28 [93]2 years ago
4 0

A type of shoot in which continuous lighting used is: 1) studio.

<h3>What is a photoshoot?</h3>

A photoshoot simply refers to a photography session which typically involves the use of digital media equipment such as a camera, to take series of pictures (photographs) of models, group, things or places, etc., especially by a professional photographer.

<h3>The types of shoot.</h3>

Basically, there are four main type of shoot and these include the following:

  • Studio
  • Underwater
  • Action
  • Landscape

In photography, a type of shoot in which continuous lighting used is studio because it enhances the photographs.

Read more on photography here: brainly.com/question/24582274

#SPJ1

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The magic square is an arrangement of numbers in a square grid in such a way that the sum of the numbers in each row, and in eac
Yuki888 [10]

Answer:

See the explanation for the answer;

Explanation:

Matlab code is as given below

-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:))      % Sum of row 1

r2= sum(M(2,:))      % Sum of row 2

r3= sum(M(3,:))      % Sum of row 3

r4= sum(M(4,:))      % Sum of row 4

r5= sum(M(5,:))      % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1))      % Sum of coloumn 1

c2= sum(M(:,2))      % Sum of coloumn 2

c2= sum(M(:,3))      % Sum of coloumn 3

c2= sum(M(:,4))      % Sum of coloumn 4

c2= sum(M(:,5))      % Sum of coloumn 5

% To find sum of diagonal

d1= sum(diag(M))               % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))      

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

   17    24     1     8    15

   23     5     7    14    16

    4     6    13    20    22

   10    12    19    21     3

   11    18    25     2     9

r1 =

   65

r2 =

   65

r3 =

   65

r4 =

   65

r5 =

   65

c1 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

d1 =

   65

d2 =

   65

3 0
3 years ago
The condensed Q formula may be used for operations in which the friction loss can be determined for:
yan [13]

The condensed Q formula may be used for operations in which the friction loss can be determined for a: 3, 4, or 5 inch hose.

<h3>What is a firehose friction loss?</h3>

A firehose friction loss can be defined as a measure of the effect of the resistance of water against the inner side of a firehose, which typically results in a pressure drop at the terminal end.

Generally, some of the factors that affect the resistance or friction in a firehose include:

  • Length of hose.
  • Age of hose.
  • Water flow (gpm)
  • Water turbulence
  • Gravity

Mathematically, the firehose friction loss can be calculated by using this formula:

FL = C × (Q/100)² × L/100.

Read more on friction loss here: brainly.com/question/17305262

#SPJ1

3 0
2 years ago
A group of friends regularly enjoys white-water rafting, and they bring piston water guns to shoot water from one raft to anothe
nydimaria [60]

Answer:

Find attached the solution

Explanation:

6 0
3 years ago
Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and
ra1l [238]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data :

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

Determine final temperature for an Isentropic  process

T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}  ----------- ( 1 )

T2 = 303 ( \frac{450}{90})^{(1.667- 1)/1.667} =  576.9K

Work done in a piston-cylinder device can be calculated using this formula

w_{in} = c_{v} ( T2 - T1 )    ------- ( 2 )

where : cv = 3.1156 kJ/kg.k  for helium gas

             T2 = 576.9K ,    T1 = 303 K

substitute given values Back to equation 2

w_{in}  = 853.4 kJ/kg

work done in a steady flow compressor can be calculated using this

w_{in} = c_{p} ( T2 - T1 )

where : cp ( constant pressure of helium gas )  = 5.1926 kJ/kg.K

             T2 = 576.9 k , T1 = 303 K

substitute values back to equation 3

w_{in} = 1422.3 kJ / kg

4 0
3 years ago
I have five brainliest why is it only showing 2?
svp [43]

Answer: Either your computer is malfunctioning or it is glitched and still thinks you only have 2. Also good job

Explanation: Leave a brainliest it helps

4 0
3 years ago
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