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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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When the outside temperature is 5.2 ⁰C, a steel beam of cross-sectional area 52 cm2 is installed in a building with the ends of
il63 [147K]

Multiply the coefficient by the change in temperature:

1.1*10^-5 x (37-5.2) = 0.0003498

Multiply Young's modulus by the area by the above answer:

2*10^11 x 52 * 0.0003498 x (1/100)^2 = 3.63792 x 10^5 N

6 0
3 years ago
Whats the purpose of the keyway
Nata [24]

Answer:

abrir candados y abrir puertas

Explanation:

4 0
2 years ago
Can I get an answer to this question please
crimeas [40]

Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

  (iii) 1,152 J

  (iv) 18Ω — maximum power transfer theorem

Explanation:

<h3>(i)</h3>

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

  (36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

<h3>(ii)</h3>

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

<h3>(iii)</h3>

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

__

<h3>(iv)</h3>

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0
2 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration
Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0
3 years ago
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
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