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Vlada [557]
3 years ago
5

Give an example of a model code.

Engineering
1 answer:
Zanzabum3 years ago
6 0

Answer:

Model codes are the codes which used to change the location of a machine from one place to other place to perform specific task.g and M are the model codes which used in machines.

Ex:

 G0 ,G1 , G2 ,G80 ,G81 ...              For motion      

  G17 ,G18 , G19                            For planer selection

 Mo ,M1 , M2 ,M30 ,M60                For Stopping

 M6                                                 For tool change

M26 ,27                                          For clamping

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An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
N76 [4]

Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

Explanation:

4 0
3 years ago
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The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in
amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be Q\frac{gallons}{minute}

Now replacing the gallon and the minute by the above values we get

Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}

Thus Q'=0.631Q\frac{liters}{second}

Now since we know that 1 liter = 0.0353ft^{3}

Using this in above relation we get

Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0
3 years ago
Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I
Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0
3 years ago
Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Un
Bogdan [553]

Answer:

Answer for the question:

Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Under which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?

Is given in the attachment.

Explanation:

5 0
3 years ago
Read 2 more answers
A steam power plant operates on the reheat Rankine cycle. Steam enters the highpressure turbine at 12.5 MPa and 550°C at a rate
gayaneshka [121]

Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg,  S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = \frac{h3-h4}{h3-h4s} = \frac{3476.5 - h4}{3476.5 - 2948.1}

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg,  s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95  and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

c) thermal efficiency

thermal efficiency = 1 - \frac{qout}{qin} = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%

8 0
3 years ago
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