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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 ( "[/]")

You might be interested in
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
What happens when the arms of the milky move away from the center of the galaxy
Alina [70]
Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes
6 0
3 years ago
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible
Zolol [24]

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note:  space time (t) =

t = \frac{A*L}{Vo}   Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = \frac{25*3*10^{-2} }{0.03} = 25 s

since the reaction is in first order

X = 1 - e^{-kt}

e^{-kt} = 1 - X

kt = In \frac{1}{1-X}

k = In \frac{1}{1-X} / t  

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then  

K = 0.156 s^{-1}

Calculating Da for a closed vessel

; Da = tk

      = 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = \frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})

therefore

\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution

3 0
3 years ago
A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com
emmainna [20.7K]

Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

3) Work, done on the system, should be calculated using the following equation: W=\int\limits^{Vb}_{Va} {p} \, dV

4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

5) Now we can calculate the integral, given in the step 3: W=mRT ln(\frac{Vb}{Va}). As we have pressure as a known values, we can re-write the equation, using pressures: W=mRT ln(\frac{pa}{pb})=1.9*0.287*279*ln(\frac{120}{160})=-260.66 kJ

Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.

6 0
3 years ago
The area under the moment diagram is shear force. a)-True b)-False
liberstina [14]

Answer:

False

Explanation:

as we know that V(x)=\frac{dM}{dx}\\ \\=> M(x) = \int\limits^x_o {V(x)} \, dx \\\\

=> Area under shear diagram gives the moment at any point but the reverse cannot be established from the same relation

8 0
3 years ago
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