Answer:
a) 
(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{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B24%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
(Ω-m)^{-1}[/tex]
b)
![= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B57%20%20%5Ctimes%2010%5E%7B-3%7D%7D%7B12.2%20%5Ctimes%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
Resistance = 121.4 Ω
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.
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.
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
Carrier frequency 100kHz
Upper side band
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
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
Answer:
Detailed step wise solution is given below:
