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fenix001 [56]
3 years ago
6

Which quantity does a charge-coupled device measure?

Engineering
1 answer:
NikAS [45]3 years ago
7 0

Explanation:

A charge-coupled device (CCD) is an integrated circuit containing an array of linked, or coupled, capacitors. Under the control of an external circuit, each capacitor can transfer its electric charge to a neighboring capacitor. CCD sensors are a major technology used in digital imaging.

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Which clauses in the Software Engineering Code of Ethics are upheld by a whistleblower (check all that apply). a. "Respect confi
garri49 [273]

Answer:

c. and d

Explanation:

As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,

In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.

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3 years ago
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kap26 [50]

Explanation:

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3 years ago
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A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:  

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:  

Step1

Reversible efficiency is calculated as follows:

\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}

0.6=\frac{0.35}{\eta_{rev}}

\eta_{rev}=0.5834

Step2

Source temperature is calculated as follows:

\eta_{rev}=1-\frac{T_{L}}{T}

\eta_{rev}=1-\frac{520}{T}

0.5834=1-\frac{520}{T}

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

6 0
3 years ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

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