3. Aqueous cleaners are

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

kiruha [24]

Answer:

percentage change in volume = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × $10^{11}$ N/m²

bulk stress = $10^{7}$ N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = $\frac{bulk\ stress}{bulk\ strain}$ ...............1

put here value and we get

3.5 × $10^{11}$ = $\frac{10^7}{bulk\ strain}$

solve it we get

bulk strain = 2.8571 × $10^{-5}$

and

bulk strain = $\frac{change\ volume}{original\ volume}$

so that percentage change in volume is = 2.8571 × $10^{-5}$ × 100

percentage change in volume = 0.00285 %

6 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room

Serggg [28]

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

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

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

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

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();

}

3 0

3 years ago

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

$h_{in} = 3353.1\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mixture)

$h_{out} = 890.1\,\frac{kJ}{kg}$

The power produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})$

$\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )$

$\dot W_{out} = 3863.98\,kW$

8 0

3 years ago

One of the most important parts about creating great photographs starts with choosing what to photograph

Sliva [168]

Yeah that is important

7 0

3 years ago