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Luba_88 [7]
3 years ago
7

A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al

l five bits are zero if the flow rate is less than 10 gallons per minute. The first bit is 1 if the flow rate is at least 10 gallons per minute; the first and second bits are 1 if the flow rate is at least 20 gallons per minute; the first, second, and third bits are 1 if the flow rate is at least 30 gallons per minute; and so on. The five bits, represented by the logical variables A, B, C, D, and E, are used as inputs to a device that provides two outputs Y and Z.
a. Write an equation for the output Y if we want Yto be 1 iff the flow rate is less than 30 gallons per minute.
b. Write an equation for the output Z if we want Z to be 1 iff the flow rate is at least 20 gallons per minute but less than 50 gallons per minute.

Engineering
1 answer:
marysya [2.9K]3 years ago
8 0

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

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Write a single statement to print: user_word,user_number. Note that there is no space between the comma and user_number. Sample
kykrilka [37]

Answer:

cout<<"''<<user_word<<"' "<<user_number;

Explanation:

The above question was answered using C++ programming language.

The keyword cout represents print and it carries out print operation only.

It prints all variable in front of it.

Assume the values of user_word and user_number to be Charles and 20, respectively.

The output of the above instruction would be

'Charles' 20 just as it is in the sample output in the question.

In java programming language, it is

System.out.print("'"+user_word+"' "+user_number);

In Qbasic, it is

PRINT "'"+user_word+"' "+ user_number

8 0
3 years ago
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc
Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m

b) f = 1 GHz = 10⁹ Hz.

\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10}  e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10}  e=-1.5 \\\\z=\frac{-1.5}{log_{10}  e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm

4 0
3 years ago
The voltage valve at which a zirconia O2S switches from rich to lean and lean to rich is
frez [133]
I think the answer is C) 0.25v I’m not sure tho
6 0
3 years ago
A design team is working on creating a new locker organizer. They have
astraxan [27]
A. Present a Solution
8 0
4 years ago
In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict
allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0
3 years ago
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