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3241004551 [841]
3 years ago
14

Global Courier Services will ship your package based on how much it weighs and how far you are sending the package. Packages abo

ve 100 pounds will not be shipped. You need to write a program in C that calculates the shipping charge.
Engineering
1 answer:
denis23 [38]3 years ago
8 0

Answer:

The code will be:

#include <stdio.h>

#include <stdlib.h>

main () {

double weight, shippingCharge, rate, segments;

int distance;

printf("Enter the weight: \n");

scanf("%lf", &weight);

printf("Enter the distance: \n");

scanf("%i", &distance);

if (weight <= 10) {

printf("Rate is $3.00 \n");

rate = 3;

} else {

printf("Rate is $5.00 \n");

rate = 5;

}

if (distance % 500 == 0) {

segments = distance / 500;

} else {

segments = distance / 500 + 1;

}

shippingCharge = rate * segments;

if (distance >1000) {

shippingCharge = shippingCharge + 10;

}

printf("Your shipping charge is $%lf\n", shippingCharge);

system ("pause");

}

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Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

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Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

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The perpendicular distance from Y axis of centroid of the mailbox

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The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4  = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

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Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

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x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

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x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

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E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

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