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Alona [7]
3 years ago
8

According to the model rules section 240.15, rules of professional conduct, licensed professional engineers are obligated to:___

___
A. ensure that design docuiments and suirveys are reviewed by a panel of licensed engineers prior to affixing a seal of approval
B. express public opinions under the direction of an employer or client regardless of knowledge of subject matter
C. practice by perfonning services only in the areas of their competence and in accordance with the cuurent standards of technical
D. offer. give or solicit services directly or indirectly in order to secure work or other valuable or political considerations
Engineering
1 answer:
Sedaia [141]3 years ago
7 0

Answer:

Option C (practice by..............technical) would be the appropriate choice.

Explanation:

  • A prerogative seems to be the discipline or nature of the design process to ensure.
  • All certificates must, therefore, throughout compliance with current requirements including professional competence, express one's right and privilege of providing offers perhaps throughout the areas of qualified professionals.

Some other alternatives that are given aren't connected to a particular scenario. But that's the best solution above.

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An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs
Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h=110\times \frac{5}{18}=30.56 m/s

Where 1 km/h=5/18 m/s

Height , h_2=40 m

g=9.81 m/s^2

Final speed, v=0

Change in kinetic energy,\Delta K.E=\frac{1}{2}m(v^2-u^2)

\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J

\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ

Where 1 KJ=1000 J

Change in potential energy,\Delta P.E=mgh(h_2-h_1)

Initially height, h1=0

Using the formula

\Delta P.E=1100\times 9.81(40-0)

\Delta P.E=431640J

\Delta P.E=431.64KJ

6 0
3 years ago
In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict
allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0
3 years ago
Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 8 MPa, 560°C and the turbine exit pressur
hoa [83]

Answer:

1. The net power developed=9370.773KW

2. Thermal Efficiency= 0.058

Explanation

Check attachment

5 0
3 years ago
Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,
solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

8 0
3 years ago
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
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