Answer:
Hello your question is poorly written attached below is the complete question
answer :
1) 60 kHz
2) Tmax = ( 1 / 34000 ) secs
Explanation:
1) Determine the Nyquist sampling frequency, (in kHz), for the input signal.
F(s) = 2 * Fmax
Fmax = 30 kHz ( since Xa(t) is band limited to 30 kHz )
∴ Nyquist sampling frequency ( F(s) ) = 2 * 30 = 60 kHz
2) Determine the largest sampling period (in s) .
Nyquist sampling period = 1 / Fs = ( 1 / 60000 ) s
but there is some aliasing of the input signal ( minimum aliasing frequency > cutoff frequency of filter ) hence we will use the relationship below
= 2π - 2π * T * 30kHz ≥ 2π * T * 4kHz
∴ T ≤ [tex]\frac{1}{34kHz}[/tex]
largest sampling period ( Tmax ) = ( 1 / 34000 ) secs
In addition to being good problem solvers, which of the following do engineers need to be?
O wealthy
O rigid
O respected
O practical
What is your favorite Electronic company? (E.g. Windows, Apple, Samsung...)
Answer:
apple
Explanation:
is the answer toWhat is your favorite Electronic company? (E.g. Windows, Apple, Samsung...)
Trichloroethylene (TCE) is a common groundwater contaminant. Based on an adult ingesting the water under residential exposure parameters, which of the following presents the greatest cancer risk:
a. to drink unchlorinated groundwater with 10 ppb of TCE
b. to switch to a surface water supply that, as a result of chlorination, has a chloroform concentration of 50 ppb
Answer:
To drink unchlorinated groundwater with 10 ppb of TCE ( A )
Explanation:
The option that presents the greatest cancer risk for ingesting water
contaminated with Trichloroethylene under the residential exposure parameters is to drink unchlorinated groundwater with 10 ppb of TCE
This is because suitable water for drinking has chloroform concentration that ranges from 4 to 44 ppb but drinking under groundwater with ppb value above 4 ppb will have a more severe damage to the body
Question 1: What is the power observed in the energy analyzer when the rated voltage(U1) is applied to the primary of the transformer, and there is no load at the secondary?
Question 2: Find the transformation ratio of the transformer using the values U1,U2 recorded in the experiment.
Question 3: Sketch the no-load operation graph of the transformer using the values U1, I2 and the values read in the energy analyzer.
Question 4: How can we find the number of turns of transformer?
Question 5: Explain the operation principle of the transformer.
Question 6: State your final observations about the experiment.
Answer:
preguntas a parte o no???????
Yeah order for a firm voltage dividers to operate properly The load resistance value should be at least Times greater than resistance value of the voltage divider bleeder resistor
Answer:
A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors.
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Reporters and Correspondents? Check all that apply.
Answer:
Acef
Explanation:
Edginuity 2021
Answer:
2,3,4,5
Explanation:
guy above me is wrong
For an Na+—Cl- ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to EA = -1.436/r ER =(7.32 *10-6 )/r8 For these expressions, energies are expressed in electron volts per Na+—Cl- pair, and r is the distance in nanometers. The net energy EN is just the sum of the preceding two expressions
Answer:
Explanation:
[tex]\text{The curve of the plot}[/tex] [tex]\mathbf{E_A,E_R, \ and \ E_N}[/tex] [tex]\text{can be seen in the attached diagram below}[/tex]
[tex]\text{From the plot}[/tex], [tex]\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}[/tex]
[tex]\mathbf{We \ knew \ that: E_N = E_A + E_R}[/tex]
[tex]\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }[/tex]
[tex]\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}[/tex]
[tex]\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}[/tex]
[tex]E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}[/tex]
[tex]\mathbf{E_o = -5.32 \ eV}[/tex]