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2 years ago

What kind of datatype is being input for Number1, Number2, Number3?

Engineering

1 answer:

Contact [7]2 years ago

6 0

Answer: integer

Explanation:

Its python code.

The datatype for number1, number2 and number3 it will be integer and even the output will be integer.

the programmer will decide whether to convert integer to float or not.

The program want a user to inputting a value number1, number2 and number3, then the interpreter will add the numbers and get the total and then the average will be total of numbers / 3.

so it will print number1, number3, number3, the sum and average.

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0

2 years ago

A system consisting of 3 lb of water vapor in a piston–cylinder assembly, initially at 350°F and a volume of 71.7 ft3, is expand

Alla [95]

Answer:

isobaric expansion = 281.09 Btu

isothermal compression= 72 Btu

Explanation:

The first law of thermodynamics is:

$Q_{AB}=W_{AB}+deltaU_{AB}$

where:

Q=heat transferred

W= work

U=internal energy

$W_{AB}=P*(V_{B}-V_{A})$

$U_{AB}=n*C_{v}(T_{2}-T_{1})$

P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.

In a isobaric process heat transferred is:

$Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})$

For an isothermal process (T2-T1 = 0) so

$Q=P*(V_{B}-V_{A})= W_{AB}$

From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.

For the first process

$Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})$

we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)

$specific-volume=\frac{71.7 ft^{3}}{3Lb}=23.9\frac{ft^{3}}{Lb}$

we use this information in the appropriate unit to find the pressure in thermodynamic tables.

T1= 176°C

v1= 1.49 m^3/kg

P=1.37 bar

in the second state we have

P=1.37 bar =137000Pa

$v_{2}=\frac{85.38ft^{3}}{3Lb}= 28.46\frac{ft^{3}}{Lb}$

with thee properties we check in the thermodynamic tables

T2= 255°C

$n=mass/Mw = 3Lb*\frac {1kg}{2.2Lb}*\frac{1000gr}{1kg}*\frac{1mol}{18gr}=75.75 mol$

we usually find Cp on tables for water but from the Mayer relation we have:

$C_{v}=C_{p}+R$

Cp for water vapor is: 33.12 J/mol*K

R=8.314 J/mol*K

Cv= 41.434 J/mol*K

replacing in the equation for Q

$Q=137000 Pa*(2.41m^{3}-2.030m^{3})+75.75mol*41.434\frac{ J}{mol*K}*(528.15-449.81 K)=296569J$

296569J =281.09 Btu

5 0

2 years ago

what is the advantage of decreasing the field current of a separately excited dc motor below its nominal value

enyata [817]

Answer:

Ability to rotate at higher speeds

Explanation:

Constant K1 becomes greater than the other constant K2

This translates to that the motor being able to rotate at high speeds, without necessarily exceeding the nominal armature voltage.

The armature voltage is the voltage that is developed around the terminals of the armature winding of an Alternating Current, i.e AC or a Direct Current, i.e DC machine during the period in which it tries to generate power.

6 0

2 years ago

simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The

Roman55 [17]

Answer:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Explanation:

Given that a simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The transmission line is 100 km copper wire of 3 cm diameter. If the voltage at the load side is 400 V,

Let first calculate the resistance in the wire.

The resistivity (rho) of a copper wire is 1.673×10^-8 ohm metres

Resistance R =( L× rho)/A

Where Area = πr^2 = π × 0.015^2

Area = 0.00071 m^2

R = (100000 × 1.673×10^-8) / 0.00071

Resistance in wire = 2.367 ohms

Then let calculate the resistance in the load.

Also, since Power P = V^2 /R

Make R the subject of formula

R = V^2/ P

R = 400^2/100000

Resistance in load = 1.6 Ohms

Current l = V / R

I = 400/1.6 = 250 Ampere

a.) Voltage drop across the line V line will be achieved by using Ohms law.

V = I R

V = 250 × 2.367

V = 591.7 v

B.) Voltage at the source side Vsource will be

V = V line + V load

V = 400 + 591.7

V = 991.7 v

C.) Percentage of the voltage drop Vline /Vsource

591.7/991.7 × 100 = 59.7%

D.) Line losses

P = I V

P = 250 × 591.7

P = 147925 W

Power loss = 147925 - 100000

Power loss = 47,925 W

Power loss = 47.9 Kw

E.) Power delivered by the source

P = IV

P = 250 × 991.7 = 247925 W

F.) System efficiency

Efficiency = power line / power source × 100

Efficiency = 147925 / 247925 × 100

Efficiency = 59.7 %

5 0

2 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac

torisob [31]

Answer:

B, its the only valid answer

5 0

2 years ago