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fiasKO [112]
3 years ago
8

Which option identifies the next step in the following scenario?

Engineering
1 answer:
Whitepunk [10]3 years ago
4 0

Answer: The engineer will create a detailed sketch that labels all of the visual components.

Explanation:

It should be noted that the reverse engineering is required for the replacement and the modification of an existing product.

With regards to the question, the correct answer is option A "The engineer will create a detailed sketch that labels all of the visual components".

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(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
Administrative controls can be broken down into two categories; programs such as
Anastaziya [24]

Administrative controls can be broken down into two categories; programs such as  HAZCOM and activities such as housekeeping is a TRUE statement.

Explanation:

  • Administrative controls (or work practice controls) are changes in work procedures such as written safety policies, rules, supervision, schedules, and training with the goal of reducing the duration, frequency, and severity of exposure to hazardous chemicals or situations.
  • Some common examples of administrative controls include work practice controls such as prohibiting mouth pipetting and recapping of needles, as well as rotating worker shifts in coal mines to prevent hearing loss.
  • Administrative controls are fourth in larger hierarchy of hazard controls, which ranks the effectiveness and efficiency of hazard controls.
  • The two categories of administrative controls programs and activities.

3 0
4 years ago
What is the condition for sampling frequency to reconstruct the information signal ?​
bagirrra123 [75]

A continuous time-varying 1-D signal is sampled by narrow sampling pulses at a regular rate fr = 1/T, which must be at least twice the bandwidth of the signal. At first, it may be somewhat surprising that the original waveform can be reconstructed exactly from a set of discrete samples.
8 0
3 years ago
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator
MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\

Carrier frequency 100kHz

Upper side band

100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

100k, 99.9k, 99.8k\ \ and \ \99.6kHz

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz

3 0
3 years ago
A powerful contribution of the 1st and 2nd laws is to determine the design feasibility of a process/device/system. Please determ
Sladkaya [172]

Answer:

Detailed step wise solution is given below:

6 0
3 years ago
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