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maksim [4K]
3 years ago
15

You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the

circuit?
Engineering
1 answer:
siniylev [52]3 years ago
5 0

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase
ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

m_{Fe} = 97.5\,g

If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

The volume percent of graphite is 91.906 per cent.

6 0
3 years ago
A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

 ( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0
3 years ago
Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more ele
shutvik [7]

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

8 0
3 years ago
In Illinois, once a person has obtained their boating education certificate what is the minimum age to operate a motorized vesse
bixtya [17]

Answer:

I dont really know, I am sorry, but I am going to ask my teacher

5 0
3 years ago
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