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Paraphin [41]
2 years ago
12

Technician A says high resistance causes an increase in current flow. Technician B says a higher than

Engineering
1 answer:
Kryger [21]2 years ago
8 0

The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.

<h3>Ohm's law</h3>

Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as

V = IR

Where

V is the potential difference

I is the current

R is the resistance

<h3>Technician A</h3>

High resistance causes an increase in current flow

V = IR

Divide both side by I

R = V / I

Thus, technician A is wrong as high resistance suggest low current flow

<h3>Technician B</h3>

Higher than normal voltage drop could indicate high resistance

V = IR

Thus, technician B is correct as high voltage indicates high resistance

<h3>Conclusion </h3>

From the above illustration, we can see that technician B is correct

Learn more about Ohm's law:

brainly.com/question/796939

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John Smeatom, U.K. 18th century, was the first self-proclaimed, civil engineer in the 18th century and IS considered “the father of modern, civil engineering”.

hoped this helped! :)

4 0
3 years ago
Read 2 more answers
3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0
3 years ago
How should you move your board through the planer? (Pick two choices.)
iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0
3 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
The best grade of hardwood lumber that is generally available is _____​
Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

5 0
1 year ago
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