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3 years ago

Two Technicians are discussing ShopKey Pro. Technician

Engineering

1 answer:

Leno4ka [110]3 years ago

6 0

Answer:

Technician B

Explanation:

i took he test already

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An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of

True [87]

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Voltage amplifier needs high input and low output resistance.

Current amplifier needs Low Input and High Output resistance.

Trans-conductance amplifier Low Input and High Output resistance.

Trans-Resistance amplifier requires High Input and Low output resistance.

Therefore, the correct answer is "None of these "

3 0

3 years ago

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

Nata [24]

<u>Explanation</u>:

For series

$\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)$

$$I=I_{1}=I_{2}=I_{n}$ (current is the same)$

$V=I R(\text {voltage is directly proportional to } R)$

$R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }$

For parallel

$\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })$

$I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})$

$I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)$

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$

3 0

3 years ago

svp [43]

Answer:

серйозних порушень точності,

∵∴⊕∴∵

5 0

3 years ago

When a starter motor begins to turn it produces a high…. *

Georgia [21]

Answer:

The motor produces a high Torque when it begins to start.

8 0

2 years ago

An engine cylinder has a stroke of 320mm and bore 280MM. Calculate the mass of air contained in the cylinder if it is filled wit

belka [17]

Answer:

The mass of the air is 0.0243 kg.

Explanation:

Step1

Given:

Stroke of the cylinder is 320 mm.

Bore of the cylinder is 280 mm.

Pressure of the air is 101.3 kpa.

Temperature of the air is 13°C.

Step2

Calculation:

Stroke volume of the cylinder is calculated as follows:

$V=\frac{\pi }{4}d^{2}L$

$V=\frac{\pi}{4}\times(\frac{280}{1000})^{2}\times(\frac{320}{1000})$

V = 0.0197 m³.

Step3

Assume air an ideal gas with gas constant 287 j/kgK. Then apply ideal gas equation for mass of the air as follows:

PV=mRT

$m=\frac{PV}{RT}$

$m=\frac{101.3\times 1000\times 0.0197}{287\times (13+273)}$

m= 0.0243 kg.

Thus, the mass of the air is 0.0243 kg.

4 0

3 years ago