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nikitadnepr [17]
3 years ago
11

The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o

f carbon dioxide c.electricity storage systems readily available d.easy to install and expand as needed epetitive costs for newer cells
Engineering
1 answer:
Xelga [282]3 years ago
6 0

Answer:

C

Explanation:

One of the disadvantages of solar cells is that electricity storage systems are not readily available. Excess energy generated by the solar panels are wasted except they are stored by solar batteries for later use. There are various systems for storing electricity from solar cells apart from solar batteries which is the common storage system. An example of another electricity storage system for solar cell is using the water electrolyzer to store solar energy which can be used to later generate hydroelectricity.

Advantages of a solar cell includes Renewable energy, Economy-friendly and environmental-friendly energy and good durability

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Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and
lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is \frac{1}{x[n] . x[n-1] x[n+1]}

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2\pi , y[n] = cos(2\pi) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0
2 years ago
When handling chemicals and solvents, technicians are recommended to
Luda [366]

Answer:

  1. To wear PPE
  2. Have prior knowledge of explosive levels and elemental properties
  3. Know procedure to eliminate any threat
7 0
3 years ago
Read 2 more answers
How do you solve this. I dont know how so I need steps if you dont mind
galben [10]

Explanation:

all I know is every number that have a bar on is equal to one

4 0
2 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
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
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