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mestny [16]
3 years ago
6

Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o

f this silicon piece? For an applied voltage of 5 V, how much current would flow though it?
Engineering
1 answer:
lukranit [14]3 years ago
7 0

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω  cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

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Read 2 more answers
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;

}

// 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

if (input == 'D')

{

// Prompt the user to input amount.

printf("Amount: ");

// input amount.

scanf("%f", &deposit_amount);

// Call to the function.

account_balance=deposit(account_balance,deposit_amount);

}

// Check if the input is W

if (input == 'W')

{

printf("Amount: ");

scanf("%f", &withdrawl_amount);

// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

}

// Check if the input is F

if (input == 'F')

{

// Dispplay the amount.

printf("Your ending balance is $%.2f\n", account_balance);

printf("Program is ending\n");

}

// End the while loop

} while(input != 'F');

return 0;

}

the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

5 0
3 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
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