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Varvara68 [4.7K]
2 years ago
13

FD=CD*((P*(V^2)*A)/2)Please solve for V

Engineering
1 answer:
REY [17]2 years ago
7 0

Answer:

The answer is V = √2FD ÷ CD × PA

Explanation:

FD = CD × PV²A ÷ 2

V² = 2FD ÷ CD × PA

V = √2FD ÷ CD × PA

Thus, The value of V is V = √2FD ÷ CD × PA

 

<u>-TheUnknownScientist 72</u>

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Modify any of the previous labs which would have crashed when non-numeric data was entered by adding exception handling so that
Mashutka [201]

Answer:

see explaination

Explanation:

import java.util.InputMismatchException;

import java.util.Scanner;

public class calculate {

static float a=0,b=0;

double cal()

{

if(a==0||b==0)

{

System.out.println("no values found in a or b");

start();

}

double x=(a*a)+(b*b);

double h=Math.sqrt(x);

a=0;

b=0;

return h;

}

float enter()

{

float val=0;

try

{

System.out.println("Enter side");

Scanner sc1 = new Scanner(System.in);

val = sc1.nextFloat();

return val;

}

catch(InputMismatchException e)

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System.out.println("Enter correct value");

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}

void start()

{

calculate c=new calculate();

while(true)

{

System.out.println("Enter Command");

Scanner sc = new Scanner(System.in);

String input = sc.nextLine();

switch(input)

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break;

case "B":

b=c.enter();

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case "C":

double res=c.cal();

System.out.println("Hypotenuse is : "+res);

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case "Q":

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public static void main(String[] args) {

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7 0
3 years ago
When a circuit has more resistance, what happens to current flow ?
zloy xaker [14]
The relationship between resistance and the area of the cross section of a wire is inversely proportional . When resistance is increased in a circuit , for example by adding more electrical components , the current decreases as a result.
6 0
3 years ago
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0
3 years ago
I want to solve the question
DedPeter [7]

Answer:

yes.

Explanation:

5 0
3 years ago
determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004
Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 ,  To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

6 0
3 years ago
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