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3 years ago

C+ Write a program that converts degrees Fahrenheit to Celsius using the following formula. degreesC = 5(degreesF – 32)/9

Engineering

1 answer:

Yanka [14]3 years ago

4 0

Answer:

Written in C++

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

float degreeC, degreeF;

cout<<"Degree Fahrenheit: ";

cin>>degreeF;

degreeC = 5 * (degreeF - 32)/9;

cout<<"Degree Celsius: "<<degreeC<<" C";

return 0;

}

Explanation:

The question requests that input should be in degree Fahrenheit

Declare all necessary variables

float degreeC, degreeF;

Prompt user for input in degrees Fahrenheit as stated in the question

cout<<"Degree Fahrenheit: ";

Get User Input

cin>>degreeF;

Convert degree Fahrenheit to Celsius

degreeC = 5 * (degreeF - 32)/9;

Display output

cout<<"Degree Celsius: "<<degreeC<<" C";

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What type of models can be communicated in more than one way.

guajiro [1.7K]

A 3-D model can be communicated, and can also be a visual model.

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4 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

$h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft$

The elevation at the high point of the road is 12186.5 in ft.

3 0

3 years ago

A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed

$T_m = 22$ because that was the room temp

$T = 31$ because the body temp when the police found it

$k=ln(\frac{8}{9})$ we solved this earlier

t = ??? we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t$

plug in the values

$t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}$

$t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}$

$t=\frac{ln(3/5)}{ln(8/9)}$

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0

3 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

3 years ago

An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg , 1980.39 MJ/Kmol

Explanation:

molar mass of ethanol (C2H5OH) = 46 g/mol

molar mass of octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol = 0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg

HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0

3 years ago