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Colt1911 [192]
4 years ago
7

What’s a pigtail when wiring

Engineering
1 answer:
Levart [38]4 years ago
7 0

Answer:

A pigtail when wiring is technique that is used in connecting a lot of wires together.

Explanation:

A pigtail is a wire that is short in length. It has two ends. One end has a connector while the other end has other wires connected to it.

Pigtail when wiring is the connection of more than one wire in a circuit to another device. Pigtail when wiring helps to extend the length of the wire in a circuit if the wire used it short).

Pigtail when wiring is a technique what helps to keep the circuit organised because it prevents the wires from getting tangled.

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50 for brainliest HELP ASAP<br> absurd answers will be recorded
s2008m [1.1K]

Answer:

1) This is because too much fuel is needed to get a payload from the surface to orbital altitude an accelerated to orbital speed.

2) This is because space travel present extreme environment that affect machines operations and survival.

Explanation:

Hope it helps

3 0
3 years ago
H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈
tiny-mole [99]

Answer:

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm  for D

v=  0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine  the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g=  acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0
3 years ago
Which of the following statements about cylinder placement are true?
Alenkinab [10]
What is the following?
3 0
3 years ago
A system consists initially of nA moles of gas A at pressure p and temperature T and nB moles of gas B separate from gas A but a
Volgvan

Answer:

A) б = - R ( nA In Ya - nB In Yb )

B) s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced

Explanation:

A) assuming ideal gas behavior the expression for entropy produced

for a closed system : s2 - s1 = б

where : s1 ( initial entropy ) = nA sA ( T, P ) + nB sB ( T, P )

s2 ( final entropy ) = nA sA ( T, YaP ) + nB sB ( T, YbP )

∴ б = - R ( nA In Ya - nB In Yb )

B) Given that

Ya and Yb are less than 1  respectively, hence the value of б  = positive

also assuming the gases are identical

s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced when same gas at same temperature and same pressure are mixed

5 0
3 years ago
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
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