If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be
- The lowest output frequency will be =
![2.5kHz](https://tex.z-dn.net/?f=2.5kHz)
<h3>
Cascade Counter</h3>
For a cascade counter,
Overall frequency = ![5*8*10*10](https://tex.z-dn.net/?f=5%2A8%2A10%2A10)
Overall frequency = ![4000](https://tex.z-dn.net/?f=4000)
<h3>Lowest F
requency</h3>
Therefore,
the lowest frequency
![F = \frac{10*10^6}{4*10^3}\\\\F = 2500Hz\\\\F = 2.5kHz](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B10%2A10%5E6%7D%7B4%2A10%5E3%7D%5C%5C%5C%5CF%20%3D%202500Hz%5C%5C%5C%5CF%20%3D%202.5kHz)
For more information on frequency, visit
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Hacking is correcttttttttt
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2
Answer:
The answer is "
"
Explanation:
Given data:
Initial temperature of tank ![T_1 = 300^{\circ}\ C= 573 K](https://tex.z-dn.net/?f=T_1%20%3D%20300%5E%7B%5Ccirc%7D%5C%20C%3D%20573%20K)
Initial pressure of tank
Diameter of throat
Mach number at exit ![M = 2.8](https://tex.z-dn.net/?f=M%20%3D%202.8)
In point a:
calculating the throat area:
![A*=\frac{\pi}{4} \times d^2](https://tex.z-dn.net/?f=A%2A%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20d%5E2)
![=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%202%5E2%5C%5C%5C%5C%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%204%5C%5C%5C%5C%3D3.14%20%5C%20cm%5E2)
Since, the Mach number at throat is approximately half the Mach number at exit.
Calculate the Mach number at throat.
![M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4](https://tex.z-dn.net/?f=M%2A%3D%5Cfrac%7BM%7D%7B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B2.8%7D%7B2%7D%5C%5C%5C%5C%3D1.4)
Calculate the exit area using isentropic flow equation.
![\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA%2A%7D%3D%20%28%5Cfrac%7B%5Cgamma%20-1%7D%7B2%7D%29%5E%7B%5Cfrac%7B%5Cgamma%20%2B1%7D%7B2%28%5Cgamma%20-1%29%7D%7D%20%20%28%5Cfrac%7B1%2B%5Cfrac%7B%5Cgamma%20-1%7D%7B2%7D%20M%2A%5E2%7D%7BM%2A%7D%29%5E%7B%5Cfrac%7B%5Cgamma%20%2B1%7D%7B2%28%5Cgamma%20-1%29%7D%7D)
Here:
is the specific heat ratio. Substitute the values in above equation.
![\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7B3.14%7D%3D%20%28%5Cfrac%7B1.4-1%7D%7B2%7D%29%5E%7B-%5Cfrac%7B1.4%2B1%7D%7B2%281.4%20-1%29%7D%7D%20%20%28%5Cfrac%7B1%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%5E2%7D%7B1.4%7D%29%5E%7B%5Cfrac%7B1.4%2B1%7D%7B2%281.4-1%29%7D%7D%20%5C%5C%5C%5CA%3D%5Cfrac%7B%5Cpi%7D%7B4%7Dd%5E2%20%5C%5C%5C%5C10.99%3D%5Cfrac%7B%5Cpi%7D%7B4%7Dd%5E2%20%5C%5C%5C%5Cd%20%3D%203.74%20%5C%20cm)
exit diameter is 3.74 cm
In point b:
Calculate the temperature at throat.
![\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K](https://tex.z-dn.net/?f=%5Cfrac%7BT%2A%7D%7BT%7D%3D%281%2B%5Cfrac%7B%5CGamma-1%7D%7B2%7D%20M%2A%5E2%29%5E%7B-1%7D%5C%5C%5C%5C%5Cfrac%7BT%2A%7D%7B573%7D%3D%281%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%5E2%29%5E%7B-1%7D%5C%5C%5C%5CT%2A%3D411.41%20%5C%20K)
Calculate the velocity at exit.
Here: R is the gas constant.
![V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}](https://tex.z-dn.net/?f=V%2A%3D1.4%20%5Ctimes%20%5Csqrt%7B1.4%20%5Ctimes%20287%20%5Ctimes%20411.41%7D%5C%5C%5C%5C%3D569.21%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D)
Calculate the density of air at inlet
![\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}](https://tex.z-dn.net/?f=%5Crho_1%20%3D%5Cfrac%7BP_1%7D%7BRT_1%7D%5C%5C%5C%5C%3D%5Cfrac%7B400%7D%7B%200.287%20%5Ctimes%20573%7D%5C%5C%5C%5C%3D2.43%5C%20%20%5Cfrac%7Bkg%7D%7Bm%5E3%7D)
Calculate the density of air at throat using isentropic flow equation.
![\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%7D%7B%5Crho_1%7D%3D%281%2B%5Cfrac%7B%5CGamma%20-1%7D%7B2%7D%20M%2A%5E2%29%5E%7B-%5Cfrac%7B1%7D%7B%5CGamma%20-1%7D%7D%20%5C%5C%5C%5C%5Cfrac%7B%5Crho%20%2A%7D%7B2.43%7D%3D%281%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%2A%5E2%29%5E%7B-%5Cfrac%7B1%7D%7B1.4-1%7D%7D%20%5C%5C%5C%5C%5Crho%2A%3D%201.045%20%5C%20%5Cfrac%7Bkg%7D%7Bm%5E3%7D)
Calculate the mass flow rate.
![m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}](https://tex.z-dn.net/?f=m%3D%20%5Crho%2A%20%5Ctimes%20A%2A%20%5Ctimes%20V%2A%5C%5C%5C%5C%3D%201.045%20%5Ctimes%203.14%20times%2010%5E%7B-4%7D%20%5Ctimes%20569.21%5C%5C%5C%5C%3D%200.186%20%5Cfrac%7Bkg%7D%7Bs%7D)
Answer:
Accuracy and precision allow us to know how much we can rely on a measuring device readings. ±.001 as a "accuracy" claim is vague because there is no unit next to the figure and the claim fits better to the definition of precision.
Explanation:
Accuracy and Precision: the golden couple.
Accuracy and precision are key elements to define if a measuring device is reliable or not for a specific task. Accuracy determines how close are the readings from the ideal/calculated values. On the other hand, precision refers to repeatability, that is to say how constant the readings of a device are when measuring the same element at different times. One of those two key concepts may not fulfill the criteria for measuring tool to be used on certain engineering projects where lack of accuracy (disntant values from real ones) or precision (not constant readings) may lead to malfunctons and severe delays on the project development.
±.001 what unit?
The manufacturer says that is an accuracy indicator, nevertheless there is now unit stated so this is not useful to see how accurate the device is. Additionally, That notation is more used to refer to device tolerances, that is to say the range of possible values the instrument may show when reading and element. It means it tells us more about the device precision during measurments than actual accuracy. I would recommend the following to the dial calipers manufacturers to better explain its measurement specifications:
- Use ±.001 as a reference for precision. It is important to add the respective unit for that figure.
- Condcut test to define the actual accuracy value an present it using one of the common used units for that: Error percentage or ppm.