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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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At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates su
UkoKoshka [18]

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

m=mass=12.5lbx1slug/32.14lb=0.39slug

F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

8 0
4 years ago
A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a l
goldfiish [28.3K]

Answer:

62.5%

Explanation:

We are given that

A ten-station transfer machine has ideal cycle time=30 sec=\frac{30}{60}=0.5 min

Frequency of line=0.075 stops per cycle

Average time=4 min

We have to determine the line efficiency

T_p=0.5+0.075(4)=0.5+0.3=0.8

Line efficiency=\frac{0.5}{0.8}\times 100=62.5%

Hence, the line efficiency=62.5%

8 0
4 years ago
Air is compressed in the compressor of a turbojet engine. Air enters the compressor at 270 K and 58 kPa and exits the compressor
tia_tia [17]

Answer:

Recall the Entropy  Balance (Second Law of Thermodynamics) equations for the system below

And for ideal gas, we know that

Change in Entropy of the air kJ/Kg.K, ΔS = S₁ - S ₂ = Cp Ln (T₂/T₁) - R Ln(P₂/P₁)

where R is gas constant =0.287kJ/kg.K and given

Cp=1.015 kJ/ kg.K

T₂ = 465k T₁=270k, P1=58kPa, P2=350kPa

substituting these values into the eqn above, we have

S = (1.015x 0.544) - (0.287x1.798)

S = 0.0363kJ/Kg.K

Hence, the mass specific entropy change associated with the compression process = 0.0363kJ/Kg.K

8 0
4 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
A car starts out from rest (zero velocity) at an elevation of 500 m and drives up a hill to reach a final elevation of 2000m and
natta225 [31]

Answer:29,930 kJ

Explanation:

Given

Car starts with an initial elevation of 500 m and drives up a hill to reach a final elevation of 2000 m

Final velocity (V)=20 m/s

Energy of car increases by 100 kJ

mass of car(m)=2000 kg

Total Energy =\Delta PE+\Delta KE+\Delta U

\Delta PE=mg(\Delta h)=2000\times 9.81\times (2000-500)

\Delta PE=29,430 kJ

\Delta KE=m\frac{v_2^2-v_1^2}{2}

\Delta KE=2000\times \frac{20^2-0^2}{2}

\Delta KE=400 kJ

\Delta U=100 kJ

Total Energy=29,430+400+100=29,930 kJ

4 0
4 years ago
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