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3 years ago

Need answers for these please

Engineering

1 answer:

Step2247 [10]3 years ago

7 0

Answer:

Following are given the answers one-by-one with explanation:

(a) 8 records

As records are the horizontal rows of data and we have 8 of them.

(b) 5 fields

As fields are vertical columns of data and we have 5 of them.

(c)E3000

As by sorting Makers_names in ascending order the last one would be Rany and by sorting Processor_type we have two option with Rany that are B1500 and E3000. We can see that E3000 will come at the end.

(d) Computer_type

As only two choices (Laptop and PC ) are being used under the column computer_type so it can be amended to contain Boolean data.

(e) format check

This will check that the field should contain one alphabet (capital letter) followed by 4 digits each.

i hope it will help you!

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all of the following are steps in the problem solving process except a. try, b. reflect, c. debug, d. define

IceJOKER [234]

Answer:

Explanation:

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3 years ago

What is the definition of insert view and why do we use it

maw [93]

Answer:

its a view point for auto cad

Explanation:

from my knowlege in IED we learned about it as a way of sing how an object would look in inventor or auto CAD

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3 years ago

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

0 - 1" -20 -15 -10 5 0 1 2 3 0

faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0

3 years ago

Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change,

Doss [256]

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

6 0

3 years ago

Need answers for these please ​

Need answers for these please