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Orlov [11]
3 years ago
11

Question 2 (Multiple Choice Worth 3 points)

Engineering
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

the car to the right

Explanation:

its in the name the RIGHT of way hope it helps good luck

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What causes decay in the amplitudes of vibration?
worty [1.4K]
Idk honestly just tryna get points.
8 0
3 years ago
In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh
astraxan [27]

( 12 17 18 19 25 )

<u>Explanation:</u>

<u>First Pass:</u>

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0
3 years ago
Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional
Anit [1.1K]

Answer:

i dont know

Explanation:

4 0
3 years ago
Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work
Naily [24]

Answer:

w_{out}=319.1\frac{BTU}{lbm}

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

s_2=s_1

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}

Best regards.

3 0
3 years ago
Using the data in the photo write the complex waveform expression​
UNO [17]

Answer:

1st Harmonic:

v(t) = 50\cos(2000\pi t)

3rd Harmonic:

v(t) = 9\cos(6000\pi t)

5th Harmonic:

v(t) = 6\cos(10000\pi t)

7th Harmonic:

v(t) = 2\cos(14000\pi t)

Explanation:

The general form to represent a complex sinusoidal waveform is given by

v(t) = A\cos(2\pi f t + \phi)

Where A is the amplitude in volts of the sinusoidal waveform

Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform

Where \phi is the phase angle in radians of the sinusoidal waveform.

1st Harmonic:

We have A = 50, f = 1000 and φ = 0

v(t) = 50\cos(2\pi 1000 t + 0) \\\\v(t) = 50\cos(2000\pi t)

3rd Harmonic:

We have A = 9, f = 3000 and φ = 0

v(t) = 9\cos(2\pi 3000 t + 0) \\\\v(t) = 9\cos(6000\pi t)

5th Harmonic:

We have A = 6, f = 5000 and φ = 0

v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)

7th Harmonic:

We have A = 2, f = 7000 and φ = 0

v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)

Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.

8 0
4 years ago
Read 2 more answers
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