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3 years ago

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A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg lid. If the local atmospheric pressure

is 101 kPa, determine the temperature at which the water starts boiling when it is heated

2 answers:

sashaice [31]3 years ago

8 0

Answer:

$T=100.2^0C$

Explanation:

The boiling point of the water depends only on the absolute pressure p. the atmospheric temperature is given by P=101000 pa and the pressure from the lid we calculate by dividing its weight with the surface area of lid. the weight is product of mass and the gravitational acceleration. surface area is calculated by given diameter.

$D=20cm=20cm.\frac{1m}{100cm} =0.2m\\A=D^2\pi /4\\p_{l} =W/A=\frac{m.g}{D^2\pi/4 }$

The absolute pressure p is sum of atmospheric pressure and the pressure from the weight of lid.

$p=p_{a} +p_{l} \\\\p=p_{a} +\frac{m.g}{D^2\pi /4} \\\\p=101000Pa+\frac{4kg.9.81m/s^2}{(0.2m^2)^2.\pi /4} \\p=102249Pa$

At this pressure the boiling temperature of water is 100.2.

Arisa [49]3 years ago

7 0

Explanation:

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Select the characteristics of an ideal operational amplifier.

SpyIntel [72]

Answer:

Numbers 4, 6, & 7 are correct

Explanation:

4- this allows the op amp to have zero voltage so that maximum voltage is transferred to output load.

6- this ensures that op amp doesn't cause loading in the original circuit, high input impedance would not deter the circuit from pulling current from it.

7- high difference between upper and lower frequencies.

3 0

3 years ago

In javaWrite a program that simulates flipping a coin to make decisions. The input is how many decisions are needed, and the out

Pavel [41]

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

import java.util.*;

public class FlipCoin

{

public static void main(String[] args)

{

// Declare Scanner

Scanner input = new Scanner (System.in);

int flips;

// Prompt to enter number of toss or flips

System.out.print("Number of Flips: ");

flips = input.nextInt();

if (flips > 0)

{

HeadsOrTails();

}

}

}

public static String HeadsOrTails(Random rand)

{

// Simulate the coin tosses.

for (int count = 0; count < flips; count++)

{

rand = new Random();

if (rand.nextInt(2) == 0) {

System.out.println("Tails"); }

else {

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rand = 0;

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7 0

3 years ago

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

3 years ago

Should you recommend changing fluids based on the color or smell of the fluid?

iragen [17]

Answer:

hi

Explanation:

Should you recommend changing fluids based on the color or smell of the fluid?

Only if the color is brown and the fluid smells burned

Yes, always.

No. Stick with the manufacturer's recommendations.

Only if the fluid is red and smells sweet

3 0

2 years ago

A magician claims that he/she has invented a novel, super-fantastic heat engine. This engine operates between two reservoirs of

Marysya12 [62]

Answer:

A) Not possible, B) Posible, C) Possible, D) Not possible.

Explanation:

The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ( $\eta$ ), no unit, is expressed below:

$\eta = 1-\frac{T_{L}}{T_{H}}$ (1)

Where:

$T_{L}$ - Cold reservoir temperature, in Kelvin.

$T_{H}$ - Hot reservoir temperature, in Kelvin.

If we know that $T_{L} = 250\,K$ and $T_{H} = 750\,K$ , then the maximum theoretical efficiency for the thermal engine is:

$\eta = 1-\frac{T_{L}}{T_{H}}$

$\eta = 0.667$

For real thermal engines, the following inequation is observed:

$0 \le \eta_{r} \le \eta$ (2)

Where $\eta_{r}$ is the efficiency of the real heat engine, no unit.

There are two possible criteria to determine if a given heat engine is real:

Efficiency

$\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}$ (3)

Where:

$Q_{L}$ - Heat rejected to the cold reservoir, in kilojoules.

$Q_{H}$ - Heat received from the hot reservoir, in kilojoules.

Power output

$W = Q_{H}-Q_{L}$ (4)

Where $W$ is the power output, in kilojoules.

Now we proceed to verify each case:

A) $Q_{H} = 900\,kJ$ , $Q_{L} = 600\,kJ$ , $W_{m} = 400\,kJ$

$\eta_{r} = 0.333$

$0 \le \eta_{r} \le \eta$

$W = 300\,kJ$

$W \ne W_{m}$

This engine is not possible.

B) $Q_{H} = 900\,kJ$ , $Q_{L} = 500\,kJ$ , $W_{m} = 400\,kJ$

$\eta_{r} = 0.444$

$0 \le \eta_{r} \le \eta$

$W = 400\,kJ$

$W = W_{m}$

The engine is possible.

C) $Q_{H} = 900\,kJ$ , $Q_{L} = 300\,kJ$ , $W_{m} = 600\,kJ$

$\eta_{r} = 0.667$

$0 \le \eta_{r} \le \eta$

$W = 600\,kJ$

$W = W_{m}$

The engine is possible.

D) $Q_{H} = 900\,kJ$ , $Q_{L} = 100\,kJ$ , $W_{m} = 800\,kJ$

$\eta_{r} = 0.889$

$\eta_{r} > \eta$

$W = 800\,kJ$

$W = W_{m}$

The engine is possible.

7 0

3 years ago