Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
Nec Article 430 covers selection of time-delay fuses for motor- overload protection.
<h3>What article in the NEC covers motor overloads?</h3>
Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .
Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.
Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.
Learn more about motor- overload from
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Answer: it’s called a saw or see saw
Explanation: it works by cutting a tree, wood, tile, etc.
According to EonCoat, corrosion is the process of decay on a material caused by a chemical reaction with its environment. Corrosion of metal occurs when an exposed surface comes in contact with a gas or liquid, and the process is accelerated by exposure to warm temperature, acids, and salts.” (1)
Although the word ‘corrosion’ is used to describe the decay of metals, all natural and man-made materials are subject to decay, and the level of pollutants in the air can speed up this process.
Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature 
= 300 F = 422.04 K
Duct temperature 
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant 
= 5.67 × 
Heat transfer per foot length is given by
Q = ∈ A ( 
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 × × 3.14 × 0.0508 × 1 × ( 
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
