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beks73 [17]
3 years ago
10

Write an if-else statement with multiple branches. If givenYear is 2101 or greater, print "Distant future (without quotes). Else

, if givenYear is 2001 or greater (2001-2100), print "21st century". Else, if givenYear is 1901 or greater (1901-2000), print 20th century". Else (1900 or earlier), print Long ago'. Do NOT end with newline. 1 #include 3 int main(void) {4 int givenYear; 6 givenYea r = 1776; 8 Your solution goes here return 0;}
Engineering
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

//Program was implemented using C++ Programming Language

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

int main()

{

int givenYear; // Declare givenYear as integer

cout<<"Enter a Given Year: "; // Prompt telling the user to enter any year

cin>>givenYear; // This line accepts input for variable givenYear

// Check if givenYear is greater than or equal to 2101

if(givenYear >= 2101)

{

cout<<"Distant Future";

}

// Check if givenYear is greater than or equal to 2001 but less than 2101

else if(givenYear >= 2001)

{

cout<<"21st Century";

}

// Check if givenYear is greater than or equal to 1901 but less than 2001

else if(givenYear >= 1901)

{

cout<<"20th Century";

}

// Check if givenYear is less than 1901

else

{

cout<<"Long ago";

}

return 0;

}

Explanation:

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6 0
4 years ago
Read 2 more answers
A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w
adelina 88 [10]

Answer:

Maximum allowable chip power is 0.35 W

Explanation:

This question is incomplete. The complete question is

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at t[infinity] = 15°c. from reliability considerations, the chip temperature must not exceed t = 85°c. f the coolant is air and the corresponding convection 200 w/m2 k, what is the maximum allowable chip power?

<u>ANSWER:</u>

The heat transfer through convection, we have the equation:

q = hA(T - T∞)

where,

q = power transfer through convection = ?

h = convection coefficient = 200 W/m²K

A = Area of convection surface = (0.005 m)² = 0.000025 m²

T = Chip surface temperature = 85° C

T∞ = Fluid temperature = 15° C

Therefore,

q = (200 W/m².K)(0.000025 m²)(85° C - 15° C)

<u>q = 0.35 W</u>

Since, difference in temperature is same on both Celsius and kelvin scale. Therefore, Celsius is written as kelvin for difference and they shall be cancelled.

3 0
4 years ago
How many gallons of water can you collect on a roof 40'×35' in a 1" rain?​
Vedmedyk [2.9K]

The answer to your question should be 630, if you have a roof the size of 40 feet, by 35 feet.

If the roof is 40 inches, by 35 inches, you would have collected 4.36 inches of rain.

Explanation: To calculate this you are going to do this:

Length of the roof ______ feet * width of roof _______ feet * .6 gallons per square feet *.75 * ________ however many inches of rain

* means multiply

7 0
4 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

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Explanation:

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