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4 years ago

How do I put a picture on here for my graph equation??

Engineering

2 answers:

Pavel [41]4 years ago

8 0

Answer:

Take a screenshot of it and when you go to question you will see a paperclip looking thing

Explanation:

Ipatiy [6.2K]4 years ago

5 0

While writing your question you should have a button at the bottom to add a picture. You should also have a button to take a picture, once you hit the button you pick/take a photo and it should add it for people to look at to help answer your question

You might be interested in

One of the dispersive components of an optical instrument is a 0.750-meter focal length monochromator equipped with the same gra

SVETLANKA909090 [29]

Answer: (a) 1.11 nm/mm

Explanation:

Monochromator is a device which is used to transmits a delectable band of wavelengths of light wavelengths available at the input.

Assuming focal length monochromator equipped with a 1200-groove/mm grating

The formula for the first-order reciprocal linear dispersion is

$D^{-1} = d/n*F$ where F is the focal length and n belongs to order of the first spectra.

$D^{-1} = \frac{(1/1200) mm * 10^6 (nm/mm)/}{1 * 0.75 m * (103 mm/m) } = 1.11 nm/mm$

Make sure you have used the conversion factors correctly as it will have a major impact on the calculation of the answer.

7 0

3 years ago

You have an ac voltage of 10 sin(60t). Design a full wave bridge rectifier which will give you an output voltage that varies a m

Masteriza [31]

As there are 10 V, for Vp1, that is the peak-voltage of the source:

$Vp1=10*\sqrt{2}=14.14 V$

Then, transformer's theory says that the relation of transformations is:

V1/V2=a

Where V1 is the voltage in the primary and V2 in the secondary.

V1=14.14 V

V2=8.55 V

a=1.65

Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:

8.5 -1.4=7.1 V

And this will be called VpL

Now we proceed to calculate the mean voltage:

$V_{mean}=VpL-(\frac{Vr}{2} )$

Where Vr is the ripple voltage, we asume that is 1 V

So, Vmean = 6.6 V

Then we have

Vmean/R= I mean

We have that R=1000 Ohm

Imedia=6.6 V/1000 Ohm

Imedia=6.6 mAmps

Finally, we can calculate the capacitor:

C=Q/Vr

C=Imean/(Vr*2f)

Where f is 60Hz

C=6.6mA/(1V*120)

C=5.5 uFarads

Therefore:

C=5.5 uFarads that works at 12 V

6 0

3 years ago

KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0

4 years ago

A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u

Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

$\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}$

where,

$\gamma$ = Specific Heat ratio of air

M = Mach number upstream

We know that $\gamma _{air}=1.4$

Applying values we get

$\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)$

Similarly the temperature downstream is obtained by the relation

$\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}$

Applying values we get

$\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C$

The Mach number downstream is obtained by the relation

$M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616$

3 0

4 years ago

This method will sell the seat in row i and column j unless it is already sold. A ticket is sold if the price of that seat in th

blsea [12.9K]

Answer:

The solution code is written in Java.

public class Movie {
private double [][] seats = new double[5][5];
private double totalSales;
public Movie(){
for(int i= 0; i < this.seats.length; i++){
for(int j = 0; j < this.seats[i].length; j++){
this.seats[i][j] = 12;
}
}
this.totalSales = 0;
}
public boolean bookSeat(int i, int j)
{
if(this.seats[i][j] != 0){
this.totalSales += this.seats[i][j];
this.seats[i][j] = 0;
return true;
}else{
return false;
}
}
}

Explanation:

The method, bookSeat(), as required by the question is presented from Line 16 - 26 as part of the public method in a class Movie. This method take row, i, and column, j, as input.

By presuming the seats is an two-dimensional array with all its elements are initialized 12 (Line 7 - 10). This means we presume the movie ticket price for all the seats are $12, for simplicity.

When the bookSeat() method is invoked, it will check if the current price of seats at row-i and column-i is 0. If not, the current price, will be added to the totalSales (Line 19) and then set the price to 0 (Line 20) and return true since the ticket is successfully sold (Line 21). If it has already been sold, return false (Line 23).

8 0

4 years ago