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2 years ago

4

you're busy and moving fast to meet your goals. you dont have a tool needed to complete your next task. a coworker is trying to

get It for you. what would you be most and least likely to do ?

1 answer:

Goryan [66]2 years ago

5 0

Let the coworker help

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Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per li

pshichka [43]

Answer:

The C code is given below with appropriate comments

Explanation:

#include<stdio.h>

//defining constants

#define DOLLAR 100

#define QUARTER 25

#define DIME 10

#define NICKEL 5

#define PENNIES 1

//converting method

void ExactChange(int userTotal,int coinVals[])

{

//checking dollars

if (userTotal >=100)

{

coinVals[0]=userTotal/DOLLAR;

userTotal=userTotal-(100*coinVals[0]);

}

//checking quarters

if (userTotal >=25)

{

coinVals[1]=userTotal/QUARTER;

userTotal=userTotal-(25*coinVals[1] );

}

//checking dimes

if (userTotal >=10)

{

coinVals[2]=userTotal/DIME;

userTotal=userTotal-(10*coinVals[2]);

}

//checking nickels

if (userTotal >=5)

{

coinVals[3]=userTotal/NICKEL;

userTotal=userTotal-(5*coinVals[3]);

}

//checking pennies

if (userTotal >=1)

{

coinVals[4]=userTotal/PENNIES;

userTotal=userTotal-coinVals[4];

}

}

//main method

int main() {

//defining the variables

int amount;

//asking for input

printf("Enter the amount in cents :");

//reading the input

scanf("%d",&amount);

//validating the input

if(amount<1)

{

//printing the message

printf("No change..!");

}

//when the input is >0

else

{

int coinVals[5]={0,0,0,0,0};

ExactChange(amount,coinVals);

//checking dollars

if (coinVals[0]>0)

{

//printing dollars

printf("%d Dollar",coinVals[0]);

if(coinVals[0]>1) printf("s");

}

//checking quarters

if (coinVals[1]>0)

{

//printing quarters

printf(" %d Quarter",coinVals[1]);

if(coinVals[1]>1) printf("s");

}

//checking dimes

if (coinVals[2]>0)

{

//printing dimes

printf(" %d Dime",coinVals[2]);

if(coinVals[2]>1) printf("s");

}

//checking nickels

if (coinVals[3]>0)

{

//prinitng nickels

printf(" %d Nickel",coinVals[3]);

if(coinVals[3]>1) printf("s");

}

//checking pennies

if (coinVals[4]>0)

{

//printing pennies

printf(" %d Penn",coinVals[4]);

if(coinVals[4]>1) printf("ies");

else printf("y");

}

}

//end of main method

}

6 0

2 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

2 years ago

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

dezoksy [38]

Answer:

(a). Entropy change of refrigerant is = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space $dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is dS = 0.0232 $\frac{KJ}{K}$

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

$T_{sat}$ = - 18.77 °c = - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant is given by

$dS_{ref} = \frac{Q}{T_{sat}}$

Since heat absorbed by refrigerant Q = 180 KJ

$dS = \frac{180}{254.23}$

dS = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space

$dS_{space} = - \frac{Q}{T_{space}}$

$T_{space}$ = - 10 °c = 263 K

$dS_{space} = - \frac{180}{263}$

$dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is given by

$dS = dS_{ref} + dS_{space}$

dS = 0.7077 - 0.6844

dS = 0.0232 $\frac{KJ}{K}$

This is the value of total entropy change.

4 0

2 years ago

A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

5 0

2 years ago

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin

juin [17]

Answer:

Answer for the question:

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routine that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)

is given in the attachment.

Explanation:

3 0

2 years ago