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3 years ago

Analog to digital conversion in digital electronics

1 answer:

7 0

In electronics, an analog-to-digital converter (ADC, A/D, or A-to-D) is a system that converts an analog signal, such as a sound picked up by a microphone or light entering a digital camera, into a digital signal.

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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

ipn [44]

Answer:

$\dot W = 1.667\, kW$

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

$\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW$

The power drawn by the air conditioner is:

$\dot W = \frac{\dot Q_{L}}{COP_{R}} \\\dot W = \frac {4.667\, kW}{2.8}\\\dot W = 1.667\, kW$

3 0

3 years ago

Which is the most effective way to reduce discrimination at the workplace? A. Appoint employees of a single race or religion. B.

sveticcg [70]

Answer:

C is the correct answer coz the others don't make sense

6 0

2 years ago

Read 2 more answers

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What

Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature $T_1$ = 15°C

Initial Pressure $P_1$ = 5 atm

Final Pressure $P_2$ = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

$\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$ $\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

$W_{lost} = T_{surr} *S$

where ;

$T_{surr}$ is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

$\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P$

$\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})$

$\delta \ S =1.408*10^{-3} \ J/gm.K$

$W_{lost} = T_{surr} *S$

$W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K$

$W_{lost} = 0.413 \ kJ/kg$

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0

4 years ago

Why are today’s homes more likely to have dangerous levels of carbon monoxide than older homes?

Margarita [4]

Answer:

Carbon monoxide has no odor, color or taste,so it cannot be detected by our senses. This means that dangerous concentrations of the gas can build up indoors and humans have no way to detect the problem until they become ill.

4 0

3 years ago

An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river

Nastasia [14]

Answer:

The answer is 2.715 In

Explanation:

An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river flow at the outlet of the basin has been observed to be 2.5 cfs, determine the annual precipitation (inches) in the basin

The annual precipitation in inches in the basin is 2.715 inches.

The solution and steps is explained in the attachment.

I hope i have been able to help.

3 0

3 years ago