Technician A is incorrect so that means Technician B is correct.
Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf
Answer:
V = 125.7m/min
Explanation:
Given:
L = 400 mm ≈ 0.4m
D = 150 mm ≈ 0.15m
T = 5 minutes
F = 0.30mm ≈ 0.0003m
To calculate the cutting speed, let's use the formula :
![T = \frac{pi* D * L}{V*F}](https://tex.z-dn.net/?f=%20T%20%3D%20%5Cfrac%7Bpi%2A%20D%20%2A%20L%7D%7BV%2AF%7D%20)
We are to find the speed, V. Let's make it the subject.
![V = \frac{pi* D * L}{F*T}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7Bpi%2A%20D%20%2A%20L%7D%7BF%2AT%7D%20)
Substituting values we have:
![V = \frac{pi* 0.4 * 0.15}{0.0003*5}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7Bpi%2A%200.4%20%2A%200.15%7D%7B0.0003%2A5%7D%20)
V = 125.68 m/min ≈ 125.7 m/min
Therefore, V = 125.7m/min
Answer:
hello your question lacks the required image attached to this answer is the image required
answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)