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Alecsey [184]
3 years ago
14

g Part 2: The features arrived the grammar Splitting categories and non-terminals gets out of hand fast. An alternative to the p

roliferation of non-terminals is to attach features to the terminals and non-terminals, and use these features to constrain the grammar. Consider the case example in part 1. We split pro into pro_nom and pro_acc, and we split np into np_nom and np_acc. The sequence det n could be either, so we added two rules (one for np_nom, and one for np_acc) with identical right-hand sides. Instead of splitting the symbols of the grammar, we can modify the lexical entries as follows:
Engineering
1 answer:
koban [17]3 years ago
3 0

Answer:

The split is given by including spaces in both tabs

Explanation:

The bracket notation can be used to indicate the split. Here is an example:

String [ ] parts = s. split ( "[/]")

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Answer: the answer is c

Explanation:

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3 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
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Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

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The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

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The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

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8 0
3 years ago
A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15%
allochka39001 [22]

Answer:

(a) the fractional conversion of pentane achieved in the furnace is  90% conversion

(b) the volumetric flow rates (Umin) of the feed air  is 256 x 10³ 1/m

(c) the volumetric flow rates (Lmin) of the gas leaving the condenser is 404.9  x 103 l/min

Explanation:

a)   Molecular weight of pentane = 72.15 g/mol

density of liquid pentane = 626 kg/m3

Flow rate of feed liquid nitrogen = 50.4 l/min

                                                  = 626*50.4*10-3

                                                   = 31.55 kg/min

                                                   = 31.55/72.15 kmol/min

                                                    = 0.4372 kmol/min

Pentane existng the burner = 3.175 kg/min

Fractional conversion = (31.55 - 3.175)/ 31.55

= 0.9 = 90% conversion

b)

C₂H₅ + 8O₂ ----------->   5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

                                                  = 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

15% excess air = 16.62*1.15 = 19.12 kmol/min

assuming air to be ideal gas

V = nRT / P.........(1)

V = 19.12 X 8.314 X 336 / 208.6

  = 256 m³ = 256 x 10³ 1/m

c)  Oxygen:

Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)

Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min

Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))

Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

                                                    = 1.966 kmol/min

= 44.362*103 l/min..........................(using (1))  

Nitrogen:

v = 0.79 x 19.12 x 8.314 x 275 / 100.325

   = 340.8 m³ / min = 340.8 x 10³ 1/min

Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min

= 404.9  x 103 l/min

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