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finlep [7]
3 years ago
13

I really need a good grade please help

Engineering
1 answer:
zysi [14]3 years ago
6 0

Answer:

A

Explanation:

50/50 bruv

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Input Energy ---> Output Energy
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motion ------> electrical. winds push the turbines which generate a magnetic fields which in turn, generates electricity

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What are the dimensions of the base of the pyramid?
Cloud [144]

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b

Explanation:

bcus it is

7 0
3 years ago
/* Function findBestVacation * duration: number of vacation days * prefs: prefs[k] indicates the rate specified for game k * pla
alexira [117]

Answer:

This is the C++ code for the above problem:

#include<bits/stdc++.h>

using namespace std;

int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {

if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days

return -1;

}

int funLevel = 0;

for(int i=start; i<start+duration; i++) { //this loop runs from starting point till

//start + duration to sum all the funlevel in plan.

funLevel = funLevel + prefs[plan[i]];

}

return funLevel;

}

int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {

int max = 0, index = 0, sum = 0 ;

for(int i=1; i<11; i++){ //this loop is to run through whole plan arry

sum = 0; //sum is initialized with zero for every call in plan ,

//in this case loop should run to 366,but for example it is 11

//as my size of plan array is 11

for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc

//fun from that index

sum = sum + prefs[plan[i]];

}

if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved

max = sum;

index = i;

}

}

return index;

}

int main() {

int ngames = 5;

int prefs[] = { 1,2,0,5,2 };

int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };

int start = 1;

int duration = 4;

cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;

cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;

cout << findBestVacation(4, prefs, ngames, plan) << endl;

}

The screen of the program is given below.

3 0
3 years ago
A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0
3 years ago
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