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Jet001 [13]
3 years ago
12

Explain the term electric current as used in engineering principles​

Engineering
2 answers:
vodomira [7]3 years ago
8 0

Answer:

<em>Electric current is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.</em><em>Current is directly proportional to voltage, inversely proportional to resistance. One of the most common electrical measurements you'll use is the watt, a unit of electrical power: W (Watts) = E (Volts) x I (Amperes). The quantity of electric charge is measured in coulombs.</em><em>They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.</em>

Explanation:

hey mate this is the best answer if you're studying engineering!

kondor19780726 [428]3 years ago
6 0

Answer:

<h2><u>An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume.</u></h2>

Explanation:

<h2><em><u>PLEASE</u></em><em><u> MARK</u></em><em><u> ME</u></em><em><u> BRAINLIEST</u></em><em><u> AND</u></em><em><u> FOLLOW</u></em><em><u> M</u></em><em><u> E</u></em><em><u> LOTS</u></em><em><u> OF</u></em><em><u> LOVE</u></em><em><u> FROM</u></em><em><u> MY</u></em><em><u> HEART</u></em><em><u> AND</u></em><em><u> SOUL</u></em><em><u> DARLING</u></em><em><u> TEJASWINI</u></em><em><u> SINHA</u></em><em><u> HERE</u></em><em><u> ❤️</u></em></h2>
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Water from the Earth's surface turns into a gas, or water vapor, when it is warmed and
bagirrra123 [75]

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Read 2 more answers
Integer to Float Conversion All labs must be done during lab time. Each labs worth 10 points The lab can be hand in next day wit
andrew-mc [135]

Answer:

Code explained below

Explanation:

.data

msg1: .asciiz "Please input a temperature in celsius: "

msg2: .asciiz "The temperature in Fahrenheit is: => "

num: .float 0.0

.text

main:

#print the msg1

li $v0, 4

la $a0, msg1

syscall

#read the float value from user

li $v0,6 #read float syscall value is $v0

syscall #read value stored in $f0

#formula for celsius to fahrenheit is

#(temperature(C)* 9/5)+32

#li.s means load immediate float

#copy value 9.0 to $f2

li.s $f2,9.0  

#copy value 5.0 to $f3

li.s $f3,5.0

# following instructions performs: 9/5

#div.s - division of two float numbers

#divide $f2 and f3.Result will stores in $f1

div.s $f1,$f2,$f3

#following instruction performs: temperature(C) * (9/5)

#multiple $f1 and $f0.Result stored in $f1

mul.s $f1,$f1,$f0

#copy value 32 to $f4

li.s $f4,32.0

#following instruction performs: (temperature(C) * (9/5))+32

#add $f1 and $f4.Result stores in $f1

add.s $f1,$f1,$f4

#store float from $f1 to num

s.s $f1,num

#print the msg2

li $v0, 4 #print string syscall value is 4

la $a0, msg2 #copy address of msg2 to $a0

#print the float

syscall

li $v0,2 #print float syscall value is 2

l.s $f12,num #load value in num to $f12

syscall

#terminate the program

li $v0, 10 #terminate the program syscall value is 10

syscall

4 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
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