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What is a p-n junction? Show by the diagram.

Natalija [7]

Answer:

The p-n junction is a region formed when a p -type semiconductor material is joined to an n-type semiconductor material

Explanation:

The p type semiconductor has holes as its majority charge carriers making it positively charged while the n –types has an overall negative charge. At the junction the holes move towards the electron until such a time when there is a balance in charges from both materials, which leads to the formation of the depletion zone as shown in the attachment below

5 0

3 years ago

The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu

marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0

3 years ago

Question 8(Multiple Choice Worth 2 points)

8090 [49]

Answer:

4 number answer is correct.

8 0

2 years ago

Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is

ddd [48]

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

cout << "\n----------------------\n";

for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

int firstDay = day-1;

//Space print

for(int i=1; i< firstDay; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

int cellCnt = 0;

// Iteration of days

for(int i=1; i<=numDays; i++)

{

//Output days

cout << left << setw(1) << "|" << setw(2) << i;

cellCnt += 1;

// New line

if ((i + firstDay-1) % 7 == 0)

{

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

cellCnt = 0;

}

// Empty cell print

if (cellCnt != 0)

{

// For printing spaces

for(int i=1; i<7-cellCnt+2; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

cout << "\n----------------------\n";

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

3 0

3 years ago

Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u

Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × $10^{-2}$ N/m

so average radius = $\frac{diameter}{2}$ = 100 µm = 100 × $10^{-6}$ m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r² .....................1

here r is radius and Δp pressure difference and σ surface tension

Δp = $\frac{2 \sigma }{r}$

put here value

Δp = $\frac{2*2.69*10^{-2}}{100*10^{-6}}$

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0

3 years ago