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:
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
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Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s 
We need to first find the parameter k from the equation below.

taking the natural logarithm of both sides:

Substituting values:

Also
 ![t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}](https://tex.z-dn.net/?f=t%5En%3D-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%5C%5Ct%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D)
Substituting values and y = 99% = 0.99
![t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s](https://tex.z-dn.net/?f=t%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D%3D%5Csqrt%5B1.7%5D%7B-%5Cfrac%7Bln%281-0.99%29%7D%7B2.76%2A10%5E%7B-4%7D%7D%7D%3D304.6s)
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion
 
        
                    
             
        
        
        
Answer:
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Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N 
                (b)  17.1 m
Explanation:  The length of wall under the surface can be given by 
                                             
The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force. 
![F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N](https://tex.z-dn.net/?f=F%28resultant%29%20%3D%20Pavg%20%28%20A%29%20%3D%20%28Patm%20%2B%20%20%5Crho%20g%20h%20c%29%2AA%20%5C%5C%3D%20%5B100000%20N%2Fm%5E2%20%2B%20%281000%20kg%2Fm%5E3%20%2A%209.81%20m%2Fs%5E2%20%2A%2025m%2F2%29%5D%2A%20%28140%2A25m%2Fsin60%29%5C%5C%3D%208.997%2A10%5E8%20N%20%5C%5C%3D%209.0%2A10%5E8%20N)
Noting from the Bernoulli  equation that 

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:
 
 
Substituting the values gives us the the distance of the surface to be equal to = 17.1 m 
 
        
             
        
        
        
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