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2 years ago

I am having trouble understanding how I got these wrong on my test. Is there something I am missing with xor?

Engineering

1 answer:

GuDViN [60]2 years ago

8 0

Answer:

your answer is correct

Explanation:

You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...

A B Y

0 0 1

0 1 0

1 0 0

1 1 1

That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.

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In your opinion, what is the external opportunity cost of a successful biking company in a community

OleMash [197]

Answer:

The cost of increased rates of accidents/injuries due to road accidents associated with cycling.

Explanation:

Opportunity cost is referred to as an alternative cost which is the value of the other choose left out while settling for a better alternative. External cost will be the cost a society has to bear as a result of a private cost. A successful biking company in a community can led to increased rates of accidents due to road accidents associated with cycling. In addition, the society will have to bear the costs of the threats presented by dangerous car drivers who posse more danger to cyclists as compared to trucks/car drivers.

8 0

2 years ago

A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

hoop stress $\sigma _{h}=\frac{Pd}{2t}$

Longitudinal stress $\sigma _{l}=\frac{Pd}{4t}$

So maximum shear tress in plane $\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}$

$\tau _{max}=\dfrac{Pd}{8t}$

Now by putting the value

$27.5=\dfrac{4\times 200}{8t}$

So t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0

2 years ago

Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.

sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh * duration

= 29700 * 6 hours ( 216000 secs )

= 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000 / 185 mi^2

= 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh / volume of runoff in equivalent depth

= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in

5 0

2 years ago

Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t

Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

%This brings an end to the program

4 0

2 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

2 years ago