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Lubov Fominskaja [6]
2 years ago
11

Why are diode logic gates not suitable for cascading operation?

Engineering
1 answer:
Vlad [161]2 years ago
5 0

Diode logic gates are not suitable for cascading operation because the pull-down or pull-up resistance within makes them a high output resistance device.

<h3>What is a cascading operation?</h3>

A cascading operation refers to a process in electrical engineering which involves the use of a circuit breaker's current-limiting capacity, so as to enable an installation of lower-rated and lower-cost circuit breakers.

According to the National Electrical Code (NEC), diode logic gates are not entirely suitable for cascading operation due to the fact that the pull-down or pull-up resistance within makes them a high output resistance device.

Read more on logic gates here: brainly.com/question/24708297

#SPJ1

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Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.

Explanation:

Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.

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3 years ago
25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an
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The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is 0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}.

Total pressure P = 1 atm = 101.325 KPa

P_{A_{1} } = 23.76 mm Hg

P_{A_{1} } = \frac{23.76}{760}

P_{A_{1} } = 0.03126 atm

P_{A_{1} } = 3.167 K Pa

When air surrounded is dry air, then PA_{2} = 0 mm Hg

R = 8.314 \frac{K P a \cdot m^{3}}{mol \cdot k}

P_{B/M} = \frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}

P_{B_{1}}=P_{T}-P_{A1_{}}

= 101.325 - 3.167

P_{B_{1} } = 98.158 K Pa

P_{B_{2}}=P_{T}-P_{A_{2}}

P_{B_{2} } = 101.325 - 0

P_{B_{2} } = 101.325 K Pa

P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}

P_{B/M} = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

N_{A} =  \frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }

N_{A} = 0.11077 × 10^{-6} mol/ m^{2}.s

N_{A} = 0.11077 × 10^{-6} × 18 × (60×60×24)

N_{A} = 0.1723 lb_{m} / m^{2}.day

\tilde{N}_{A}=N_{A} \times Area

Area of individual pipe is

A=\frac{\pi}{4}(0.0254)^{2}

A = 0.00051 m^{2}

\bar{N}_{\boldsymbol{A}} = 0.1723 × 0.00051

\bar{N}_{\boldsymbol{A}} = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

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