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BabaBlast [244]
2 years ago
5

// This pseudocode segment is intended to compute and display // the average grade of three tests start Declarations num test1 n

um test2 num test2 num average output "Enter score for test 1 input test1 output "Enter score for test 2 " input test2 output "Enter score for test 3 " input test3 average = test1 + test2 + test3 / 3 output "Average is ", answer end
Engineering
1 answer:
enot [183]2 years ago
5 0

Answer:

start

Declarations

num test1

num test2

num test2

num average

output "Enter score for test 1 "

input test1

output "Enter score for test 2 "

input test2

output "Enter score for test 3 "

input test3

average = (test1 + test2 + test3) / 3

output "Average is ", average

end

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2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
gladu [14]

Answer:

we use

sigma=force /area\\A=\pi r^2\\D=2*r

5 0
2 years ago
Python Homework
DerKrebs [107]

Answer:

  1. TAX_RATE = 0.20
  2. STANDARD_DEDUCTION = 10000.0
  3. DEPENDENT_DEDUCTION = 3000.0
  4. # Request the inputs
  5. grossIncome = float(input("Enter the gross income: "))
  6. numDependents = int(input("Enter the number of dependents: "))
  7. # Compute the income tax
  8. taxableIncome = grossIncome - STANDARD_DEDUCTION - \
  9. DEPENDENT_DEDUCTION * numDependents
  10. incomeTax = taxableIncome * TAX_RATE
  11. # Display the income tax
  12. print("The income tax is $" + str(round(incomeTax,2)))

Explanation:

We can use round function to enable the program to output number with two digits of precision.

The round function will take two inputs, which is the value intended to be rounded and the number of digits of precision. If we set 2 as second input, the round function will round the incomeTax to two decimal places. The round function has to be enclosed within the str function so that the rounded value will be converted to a string and joined with the another string to display the complete a sentence of income tax info.

5 0
2 years ago
A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected
Viktor [21]

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload)   ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

7 0
2 years ago
4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and
hodyreva [135]
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
6 0
2 years ago
An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t
11Alexandr11 [23.1K]

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

                length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

                                 ∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J =  3.29* 10^{-8} eV.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

                                                    λ = c/v

                                                    dλ/dv = -c/v² = -λ²/c

Therefore,

      ∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

                                 =  9.20 × 10⁻¹⁵ m or 9.20 fm

     The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

  = (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m          

                                                   

6 0
2 years ago
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