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lorasvet [3.4K]
1 year ago
9

The maximum capacity of a 2-lane carriageway of a four lane dual carriageway is 2000 veh/hour. due to pipe laying operations the

width of two lane carriageway is reduced, restricting the maximum capacity to 1100 veh/hour. when the flow upstream beyond the influence of the bottleneck is reasonably steady and freeflowing at 1500 veh/hour. find i) the mean speed of traffic in the bottleneck ii) the rate at which the queue of the congested conditions outside bottleneck grows. the mean space headway when the vehicles are stationary is 8 m. the relation between speed and concentration is linear
Engineering
1 answer:
Yuki888 [10]1 year ago
3 0

Based on the maximum capacity as a result of the pipe-laying operations, and the maximum capacity without obstruction to the four-lane dual carriageway, the mean speed of traffic in the bottleneck is -2.73km.

<h3>What is the mean speed of traffic?</h3>

This can be found as:

= (Maximum restricted capacity - freeflowing rate) / (Kb - Ka)

= (1,100 - 1,500) / (209 - 62.5)

= -2.73 km/h

The rate that the queue outside the bottleneck grows is:

= Free flowing rate - (mean speed of traffic x ka)

= 1,500 - (-2.73 x 62.5)

= 1,670 veh/hour

Find out more on bottleneck at brainly.com/question/9551615

#SPJ1

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Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balanc
GarryVolchara [31]

Answer:

Explanation:

Sample output:

BANK ACCOUT PROGRAM!

----------------------------------

Enter the old balance: 1234.50

Enter the transactions now.

Enter an F for the transaction type when you are finished.

Transaction Type (D=deposit, W=withdrawal, F=finished): D

Amount: 568.34

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 25.68

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 167.40

Transaction Type (D=deposit, W=withdrawal, F=finished): F

Your ending balance is $1609.76

Program is ending

Code to copy:

// include the necessary header files.

#include<stdio.h>

// Definition of the function

float withdraw(float account_balance, float withdraw_amount)

{

// Calculate the balace amount.

float balance_amount = account_balance - withdraw_amount;

// Check whether the withdraw amount

// is greater than 0 or not.

if (withdraw_amount > 0 && balance_amount >= 0)

{

// Assign value.

account_balance = balance_amount;

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// return account_balance

return account_balance;

}

// Definition of the function deposit.

float deposit(float account_balance, float deposit_amount)

{

// Check whether the deposit amount is greater than zero

if (deposit_amount > 0)

{

// Update account balance.

account_balance = account_balance + deposit_amount;

}

// return account balance.

return account_balance;

}

int main()

{

// Declare the variables.

float account_balance;

float deposit_amount;

float withdrawl_amount;

char input;

// display the statement on console.

printf("BANK ACCOUT PROGRAM!\n");

printf("----------------------------------\n");

// prompt the user to enter the old balance.

printf("Enter the old balance: ");

// Input balance

scanf("%f", &account_balance);

// Display the statement on console.

printf("Enter the transactions now.\n");

printf("Enter an F for the transaction type when you are finished.\n");

// Start the do while loop

do

{

// prompt the user to enter transaction type.

printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");

// Input type.

scanf(" %c", &input);

// Check if the input is D

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{

// Prompt the user to input amount.

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// input amount.

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// Call to the function.

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}

// Check if the input is W

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printf("Amount: ");

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// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

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// Check if the input is F

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// Dispplay the amount.

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// End the while loop

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the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

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3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

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Explanation:

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l = 1\,km

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\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

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\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

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\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

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