Answer:
<h3>
advantages: </h3>
<em>lower power consumption, modulation system is simple</em>
<h3>disadvantages<em>:</em></h3>
<em>complex detection</em>
<h3><em>applications:</em></h3>
analog TV systems: to transmit color information
<h3><em /></h3>
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Explanation:
Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
![E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\](https://tex.z-dn.net/?f=E%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%20%2B6.32%2B6.33%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20A%20%29%20%3D%206.34%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%20%2B6.31%2B6.32%2B6.35%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%206.3225%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%20%2B6.32%2B6.36%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%206.355%5C%5C)
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
![Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\](https://tex.z-dn.net/?f=Var%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%5E2%2B6.32%5E2%2B6.33%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.34%5E2%5C%5C%5C%5CVar%20%28%20A%20%29%20%3D%200.00025%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%5E2%2B6.31%5E2%2B6.32%5E2%2B6.35%5E2%7D%7B4%7D%20-%206.3225%5E2%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%200.00026875%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%5E2%2B6.32%5E2%2B6.36%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.355%5E2%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%200.000475%5C%5C)
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
![P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\](https://tex.z-dn.net/?f=P%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.34-6.321%7D%7B6.321%7D%2A100%3D%200.30058%5C%5C%5C%5CP%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.3225-6.321%7D%7B6.321%7D%2A100%3D%200.02373%5C%5C%5C%5CP%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.355-6.321%7D%7B6.321%7D%2A100%3D%200.53788%5C%5C)
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate
Answer:
λ^3 = 4.37
Explanation:
first let us to calculate the average density of the alloy
for simplicity of calculation assume a 100g alloy
80g --> Ag
20g --> Pd
ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)
= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)
= 10744.62 kg/m^3
now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate
total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27
= 1.12*10^23 unit cells
mass of Pd in 1 unit cell = 20/1.12*10^23
Now,
ρ_avg= mass of unit cell/volume of unit cell
ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3
λ^3 = 4.37
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ
![$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Cphi%20%3D%20%5Cfrac%7B%281%2C0%2C0%29%20%5Ccdot%20%281%2C1%2C0%29%7D%7B1%20%5Ctimes%20%5Csqrt2%7D%24)
![$=\frac{1}{\sqrt2 }$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B%5Csqrt2%20%7D%24)
![$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Clambda%20%3D%20%5Cfrac%7B%281%2C0%2C0%29%20%5Ccdot%20%281%2C-1%2C1%29%7D%7B1%20%5Ctimes%20%5Csqrt3%7D%24)
![$=\frac{1}{\sqrt3 }$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B%5Csqrt3%20%7D%24)
τ = σ cos Φ cos λ
∴ ![$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $](https://tex.z-dn.net/?f=%2450%3D%20%5Csigma%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt3%7D%20%24)
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]
![$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Cphi%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%20-1%2C%200%29%7D%7B1%20%5Ctimes%20%5Csqrt2%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt2%7D%24)
![$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Clambda%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%201%2C%20-1%29%7D%7B1%20%5Ctimes%20%5Csqrt3%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt3%7D%24)
τ = σ cos Φ cos λ
∴ ![$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $](https://tex.z-dn.net/?f=%2450%3D%20%5Csigma%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt3%7D%20%24)
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]
![$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Cphi%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%200%2C%201%29%7D%7B1%20%5Ctimes%20%5Csqrt2%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt2%7D%24)
![$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Clambda%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%201%2C%20-1%29%7D%7B1%20%5Ctimes%20%5Csqrt3%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt3%7D%24)
τ = σ cos Φ cos λ
∴ ![$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $](https://tex.z-dn.net/?f=%2450%3D%20%5Csigma%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt3%7D%20%24)
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]
![$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Cphi%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%200%2C%20-1%29%7D%7B1%20%5Ctimes%20%5Csqrt2%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt2%7D%24)
![$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$](https://tex.z-dn.net/?f=%24%5Ccos%20%5Clambda%20%3D%20%5Cfrac%7B%281%2C%200%2C%200%29%20%5Ccdot%20%281%2C%20-1%2C%201%29%7D%7B1%20%5Ctimes%20%5Csqrt3%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt3%7D%24)
τ = σ cos Φ cos λ
∴ ![$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $](https://tex.z-dn.net/?f=%2450%3D%20%5Csigma%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B%5Csqrt3%7D%20%24)
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
Answer:
The capacity of the sludge pump is 0.217 m3/min
Explanation:
Solution is attached below