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2 years ago

13

A 10MHz clock frequency is applied to a cascaded counter consisting of a modulus-5 counter, a modulus-8 counter and two modulus-

10 counters. The lowest output frequency available will be:

1 answer:

rewona [7]2 years ago

4 0

If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be

The lowest output frequency will be = $2.5kHz$

<h3>Cascade Counter</h3>

For a cascade counter,

Overall frequency = $5*8*10*10$

Overall frequency = $4000$

<h3>Lowest Frequency</h3>

Therefore,

the lowest frequency

$F = \frac{10*10^6}{4*10^3}\\\\F = 2500Hz\\\\F = 2.5kHz$

For more information on frequency, visit

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Wastewater flows into a _________ once it is released into a floor drain.

vodomira [7]

Answer:

a

Explanation:

5 0

3 years ago

LC3 Programming ProblemUse .BLKW to set up an array of 10 values, starting at memory location x4000, as in lab 4.Now programmati

irga5000 [103]

Answer:

Check the explanation

Explanation:

Code

.ORIG x4000

;load index

LD R1, IND

;increment R1

ADD R1, R1, #1

;store it in ind

ST R1, IND

;Loop to fill the remaining array

TEST LD R1, IND

;load 10

LD R2, NUM

;find tw0\'s complement

NOT R2, R2

ADD R2, R2, #1

;(IND-NUM)

ADD R1, R1, R2

;check (IND-NUM)>=0

BRzp GETELEM

;Get array base

LEA R0, ARRAY

;load index

LD R1, IND

;increment index

ADD R0, R0, R1

;store value in array

STR R1, R0,#0

;increment part

INCR

;Increment index

ADD R1, R1, #1

;store it in index

ST R1, IND

;go to test

BR TEST

;get the 6 in R2

;load base address

GETELEM LEA R0, ARRAY

;Set R1=0

AND R1, R1,#0

;Add R1 with 6

ADD R1, R1, #6

;Get the address

ADD R0, R0, R1

;Load the 6th element into R2

LDR R2, R0,#0

;Display array contents

PRINT

;set R1 = 0

AND R1, R1, #0

;Loop

;Get index

TOP ST R1, IND

;Load num

LD R3,NUM

;Find 2\'s complement

NOT R3, R3

ADD R3, R3,#1

;Find (IND-NUM)

ADD R1, R1,R3

;repeat until (IND-NUM)>=0

BRzp DONE

;load array address

LEA R0, ARRAY

;load index

LD R1, IND

;find address

ADD R3, R0, R1

;load value

LDR R1, R3,#0

;load 0x0030

LD R3, HEX

;convert value to hexadecimal

ADD R0, R1, R3

;display number

OUT

;GEt index

LD R1, IND

;increment index

ADD R1, R1, #1

;go to top

BR TOP

;stop

DONE HALT

;declaring variables

;set limit

NUM .FILL 10

;create array

ARRAY .BLKW 10 #0

;variable for index

IND .FILL 0

;hexadecimal value

HEX .FILL x0030

;stop

.END

7 0

3 years ago

A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat

ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n ..........1

and

total cost of capital intensive = $1,244,000 + $92.5 n ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n < $33800 + $143 n

solve we get

n > $\frac{906000}{50.5}$

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > $\frac{1244000}{104.5}$

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0

3 years ago

A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying

Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius, $r_{2}$ = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, $r_{1}$ = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

τmax $= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )$

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0

3 years ago