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.
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Answer: the increase in the external resistor will affect and decrease the current in the circuit.
Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is
E = IR + Ir = I(R + r)........(1)
Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)
I = E/(R + r)
As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.
Answer:
The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v
Explanation:
Solution
Given that:
V+ = 20v
Re = 2kΩ
Rc = 1kΩ
Now we will amke use of the method KVL in the loop.
= - Ve + IE . Re + VEB + VB = 0
Thus
IE = V+ -VEB -VB/Re
Which gives us the following:
IE = 20-0.7 - 10/2k
= 9.3/2k
so, IE = 4.65 mA
IB = IE/β +1 = 4.65 m /101
Thus,
IB = 0.046039 mA
IB = 46.039μA
IC =βIB
Now,
IC = 100 * 0.046039
IC is 4.6039 mA
Now,
VB = 10v
VE = VB + VEB
= 10 +0.7 = 10.7 v
So,
Vc =Ic . Rc = 4.6039 * 1k
=4.6039 v
Finally, this is the table summary from calculations carried out.
Summary Table
Parameters IE IC IB VE VB Vc
Unit mA mA μA V V V
Value 4.65 4.6039 46.039 10.7 10 4.6039
Answer:
There are 2 expected readings greater than 2.70 V
Solution:
As per the question:
Total no. of readings, n = 60 V
Mean of the voltage, 
standard deviation, 
Now, to find the no. of readings greater than 2.70 V, we find:
The probability of the readings less than 2.70 V, 
:
Now, from the Probability table of standard normal distribution:
Now,
Now, for the expected no. of readings greater than 2.70 V:
No. of readings expected to be greater than 2.70 V = 
No. of readings expected to be greater than 2.70 V = 
≈ 2
Answer:
The Full details of the answer is attached.
