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Anna [14]
3 years ago
13

Given the following Grammar and the right sentential form draw a parse tree and show the phrases, simple phrases and handle.S →

AbB | bAcA → Ab | aBBB → Ac | cBb | ca) baabccBbc b) ccBbcbbbB c)baBcBbbc

Engineering
2 answers:
IRISSAK [1]3 years ago
7 0

Answer:

<u>Answer is written in attached document.</u>

Download docx
Shtirlitz [24]3 years ago
4 0

Answer:

Find the answer in the attached in the order

Option a, Option c and Option b

Explanation:

You might be interested in
What is the definition of a duty cycle?
ira [324]

Answer:

D=\frac{PW}{T}*100

Explanation:

In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.

The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:

D=\frac{PW}{T}*100

Where:

D=Duty\hspace{3}Cycle

PW=Pulse\hspace{3}Active\hspace{3}Time

T=Period\hspace{3}of\hspace{3}the\hspace{3}Signal

Here is a picture that will help you understand these concepts.

5 0
3 years ago
A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig
jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

             Section: 9-ft section of pipe.

                     Sum of forces perpendicular to member AC = 0

                     F_d - 0.8*W*L = 0

                     F_d = 0.8*10*9 = 72 lb

                     Sum of forces perpendicular to member BC = 0

                     F_e - 0.6*W*L = 0

                     F_e = 0.6*10*9 = 54 lb

              F_d = 72 lb ,  F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

              Section: Entire Frame.

                    Sum of moments about point B = 0

                    -A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

                    -A_y*(1.5625) + 15 + 39.375 = 0

                    A_y = 34.8 lb  

                   Sum of forces in vertical direction = 0

                     A_y + B_y - 0.8*F_d - 0.6*F_e = 0

                     B_y = 0.8*(72) + 0.6*(54) - 34.8

                     B_y = 55.2 lb  

                   Sum of forces in horizontal direction = 0

                     A_x + B_x - 0.6*F_d + 0.8*F_e = 0

                     A_x + B_x = 0

               Section: Member AC

                    Sum of moments about point C = 0

                     F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

                     72*2.5 - 34.8*12 - 9*A_x = 0

                     A_x = -237.6 / 9 = - 26.4 lb

                     B_x = - A_x = 26.4 lb

                     A_x = -26.4 lb  ,  B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

               Member AC:

                    From point A till just before point D

                     -0.6*A_x*x - A_y*0.8*x + M = 0

                     15.84*x - 27.84*x + M = 0

                      M = 12*x   ..... max value at D, x = 12.25 in

                      M_max = 12*12.25/12 = 12.25 lb-ft

               Member BC:

                    From point B till just before point E

                     -0.8*B_x*x + B_y*0.6*x + M = 0

                     -21.12*x + 33.12*x + M = 0

                      M = -12*x   ..... max value at E, x = 11.25 - 2.5 = 8.75 in

                      M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

                      AC: at D , M_max = 12.25 lb-ft

                      BC: at E , M_max = 8.75 lb-ft

5 0
3 years ago
Safety measures to be taken during technical drawing<br>​
Pavlova-9 [17]

Explanation:

Couldn't you just leave it in centimeters?

3 0
2 years ago
At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286
deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s

calculation fro friction factor

from standard table we have

\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}

Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4

so for calculated R and\frac{\epsilon}{D}, F value = 0.0212

from equation 1

h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}

h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}

h_p = 826.76ft so that

P =  \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp

3 0
3 years ago
What is the answer to life the universe and everything <br> (worth 95 points!)
taurus [48]
I would say nothing, time is endless and the cycle of life is endless, not only on earth but almost anywhere, people try to find answers like what’s at the bottom of the ocean and stuff like that but there’s no point in finding out because it has no benefit, life is made for different reasons so there’s not one answer to it
5 0
3 years ago
Read 2 more answers
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