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3 years ago

15

Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit cont

rols current flow. Who is right?

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

A

Explanation:

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2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.

hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0

3 years ago

What is polarized electrical receptacle used for

lidiya [134]

Answer:

they are used for electrical currents so that they can flow along the appropriate wires in the circuit

Explanation:

4 0

4 years ago

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

$E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\$

E=52000Hp.h

$E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\$

E=38724920Wh

$E=52000Hph(\frac{1977378.4 ft lb}{1Hph}$

E=1.028x10^11 ftlb

3 0

3 years ago

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0

3 years ago

Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

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4 0

1 year ago