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rusak2 [61]
3 years ago
7

When you're running a test bench, you would like to include a printout to the screen to inform yourself on the status of the cur

rent simulation. Which command you could use to monitor specific variables or signals in a simulation every time one of the signals changes value?
Engineering
1 answer:
boyakko [2]3 years ago
4 0

Answer:

$Monitor

Explanation:

The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)

is the $Monitor command

This command is also used to monitor the varying values of signals during simulation.

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A parallel circuit has a resistance of 280 and an inductive reactance of 360 02. What's this circuit's impedance?
Doss [256]

Answer:

540 W

Explanation:

4 0
3 years ago
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
3 years ago
A 2-m^3 rigid tank contains nitrogen gas at 500kPa and 300K. Now the gas is compressed isothermally to a volume of 0.1 m. The wo
stira [4]

Answer:

(d) 2996 kJ

Explanation:

We have given initial volume v_1=2m^3

initial pressure p_1=500\ kPa

initial temperature T_1=300\ K

we know that during  isothermal process the work done is given by

W=p_1v_1ln\frac{v_2}{v_1}

where v_2= final\ volume =0.1m^3\ given

putting all these value in formula of work done

W=500\times 2\times ln\frac{0.1}{2}

=-2995.8 kJ here negative sign indicates that work is dine on the gas

so wok done =2995.8 kJ

8 0
3 years ago
What is the fastest motorcycle in the world ?
givi [52]

Answer:

Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...

MTT Turbine Superbike Y2K – top speed: 227 mph. This bike is one of the most powerful production motorcycles. ...

Suzuki Hayabusa – top speed: 248 mph. 1340cc

8 0
3 years ago
Read 2 more answers
Two added to four times a number, minus 3 times the number, equals 5.
vladimir1956 [14]
<h2>Answer:</h2>

<u>x= 3</u>.

<h2>Explanation:</h2>

<em>What is presented in this problem is basically an equation in verbal form.</em>

<em />

<h3>1. Write the equation.</h3>

2+4x-3x=5

<h3>2. Solve for x.</h3>

2+4x-3x=5\\ \\2+x=5\\ \\x=5-2\\ \\x=3

<h3>3. Express the result.</h3>

x= 3.

8 0
2 years ago
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