All categories

3 years ago

Convert 0.025 in into mm.

Engineering

2 answers:

Leto [7]3 years ago

8 0

Answer:

.025 inch = 0.635 mm

Explanation:

We know that 1 inch = 2.54 cm

also we know that 1 cm = 10 mm

Thus 1 inch = 2.54 x 10 mm

=> 1 inch = 25.4 mm

Hence 0.025 inch = 25.4 x .025 mm

Thus .025 inch = 0.635 mm

malfutka [58]3 years ago

3 0

Answer: 0.025 in = 0.065 mm

Explanation: To convert the value in inches to mm we have to multiply the inches by the conversion factor 25.4.

So, 0.025 × 25.4 = 0.065 mm (millimeter)

Conversion formula for calculation in (inch) into mm is:

Value in mm = Value in in × 25.4

One inch is equal to the 25.4 mm.

You might be interested in

In homes today, what is behind the reason for flashover fires occurring much more rapidly than in the past generations?

garik1379 [7]

Answer:

One of the reasons why flashover fires are more prevalent today than it was in the past is that homes and furniture today are made from materials that are far more combustible than those of previous years.

Explanation:

A flashover fire is the rapid ignition and combustion of all flammable materials in an enclosed vicinity in a very short period of time.

Thirty years ago, the average escape time from a house that was on fire is about sixteen and fifty seconds...that would be approximately seventeen minutes. Presently that figure is down to four minutes.

One of the reasons identified is that the internal and external appurtenances especially furniture in use today are more combustible than those of previous years. That is, as they burn, they produce more heat and disintegrate faster.

The reason identified for this is, old houses were made of more natural materials such as real wood etc whilst the furniture and curtains in modern houses are mostly from synthetic materials.

Cheers

6 0

3 years ago

_______ is a material property that pertains to local resistance to plastic deformation, such as scratching or denting. It is of

Readme [11.4K]

Answer: hardness

Explanation:

Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.

7 0

3 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago

Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is

GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

public static void main(String[] args) {

final int MAXCHEESE = 10;

String[] names = new String[MAXCHEESE];

double[] prices = new double[MAXCHEESE];

double[] amounts = new double[MAXCHEESE];

// Three Special Cheeses

names[0] = "Humboldt Fog";

prices[0] = 25.00;

names[1] = "Red Hawk";

prices[1] = 40.50;

names[2] = "Teleme";

prices[2] = 17.25;

System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

System.out.println(names[0] + ": $" + prices[0] + " per pound");

System.out.println(names[1] + ": $" + prices[1] + " per pound");

System.out.println(names[2] + ": $" + prices[2] + " per pound");

Random ranGen = new Random(100);

// error at initialising i

// i should be from 0 to MAXCHEESE value

for (int i = 0; i < MAXCHEESE; i++) {

names[i] = "Cheese Type " + (char) ('A' + i);

prices[i] = ranGen.nextInt(1000) / 100.0;

amounts[i] = 0;

System.out.println(names[i] + ": $" + prices[i] + " per pound");

}

7 0

3 years ago

Can someone help me plz!!

pogonyaev

It has to do with mechanical engineering

6 0

2 years ago