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2 years ago

What gadgets are charge coupled devices used in?

Engineering

1 answer:

Alinara [238K]2 years ago

5 0

Answer:

They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners

Explanation:

A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.

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ShoppingBay is an online auction service that requires several reports. Data for each auctioned item includes an ID number, item

daser333 [38]

Answer:

START

READ ID_Number

READ Item_description

READ length_of_auction_Days

READ minimum_required_bid

IF minimum_required_bid GREATER THAN 100

THEN

DISPLAY

Item Details are

Item Id : ID_Number

Item Description: Item_description

Length Action days: length_of_auction_Days

Minimum Required Bid: minimum_required_bid

END

Explanation:

5 0

3 years ago

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

3 years ago

What is the ANSI B paper size also know as?

krek1111 [17]

Answer:

ANSI A sized paper is commonly referred to as Letter and ANSI B as Ledger or Tabloid.

Explanation:

7 0

3 years ago

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

Klio2033 [76]

Answer:

A) ΔS_refrigerant = 0.70754 Kj/K

B) ΔS_space = -0.68441 Kj/K

C) ΔS_total = 0.02313 Kj/K

Explanation:

A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C

Converting to degree kelvin yields;

T = -18.77 + 273 = 255.23 K

Formula for entropy change of refrigerant is given as;

ΔS_refrigerant = Q_in/T_refrigerant

We are given Q = 180 KJ

Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K

B) Formula for entropy change of cooled space is given as;

ΔS_space = Q_out/T_s pace

T_space = -10°C = 273 - 10 = 263K

Thus, ΔS_space = -180/263 = -0.68441 Kj/K

C) the total entropy change would be;

ΔS_total = ΔS_refrigerant + ΔS_space

Thus,

ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K

8 0

3 years ago

Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an

Lera25 [3.4K]

Answer:

$\dot Q_{out} = 13369.104\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0$

The rate of heat transfer between the turbine and its surroundings is:

$\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}$

The specific enthalpies at inlet and outlet are, respectively:

$h_{in} = 3076.41\,\frac{kJ}{kg}$

$h_{out} = 2675.0\,\frac{kJ}{kg}$

The required output is:

$\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW$ $\dot Q_{out} = 13369.104\,kW$

4 0

3 years ago