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1 year ago

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Engineering

1 answer:

Flura [38]1 year ago

6 0

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

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Manufacturing employees who perform assembly line work are referred to as

mamaluj [8]

Answer:

C. assembly line workers.

Explanation:

8 0

3 years ago

Read 2 more answers

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

;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

Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m

Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

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3 years ago

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LuckyWell [14K]

Answer:

Feedback control of arm movements using Neuro-Muscular Electrical Stimulation (NMES) combined with a lockable, passive exoskeleton for gravity compensation

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2 years ago

Imagine you are a process safety consultant and you have been tasked to make a metal refinery site DSEAR compliant. What are the

masya89 [10]

Complying with DSEAR involves:

Assessing risks. ...

Preventing or controlling risks. ...

Control measures. ...

Mitigation. ...

Preparing emergency plans and procedures. ...

Providing information, instruction and training for employees. ...

Places where explosive atmospheres may occur ('ATEX' requirements)

hse uk

4 0

2 years ago