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torisob [31]
3 years ago
9

What do you think causes the differences in the properties of oxygen gas (O2) and ozone (O3)? the space between the atoms the ra

tio of atoms in the molecules cannot tell the number of oxygen atoms
Engineering
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

Oxygen is an important part of the atmosphere and is necessary to sustain terrestrial life. Because it comprises most of the mass in water, it also comprises most of the mass of living organisms. All major classes of structural molecules in living organisms, such as proteins, carbohydrates, and fats, contain oxygen, as do the major inorganic compounds that comprise animal shells, teeth, and bone. Elemental oxygen (O2) is produced by cyanobacteria, algae, and plants through the process of photosynthesis, and is used in cellular respiration by most living organisms on earth. Oxygen is toxic to obligate anaerobic organisms (organisms which need a lack of oxygen for survival), which were the dominant form of early life on Earth, until O2 began to accumulate in the atmosphere.

Explanation:

You might be interested in
What gadgets are charge coupled devices used in?
Alinara [238K]

Answer:

They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners

Explanation:

A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.

5 0
2 years ago
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0
3 years ago
Write the implementation (.cpp file) of the Player class from the previous exercise. Again, the class contains:
hoa [83]

Answer:

//Define the header file

#ifndef PLAYER_H

#define PLAYER_H

//header file.

#include <string>

//Use the standard namespace.

using namespace std;

//Define the class Player.

class Player

{

//Declare the required data members.

string name;

int score;

public:

//Declare the required

//member functions.

void setName(string par_name);

void setScore(int par_score);

string getName();

int getScore();    

}

//End the definition

//of the header file.

#endif

Player.cpp:

//Include the "Player.h" header file,

#include "Player.h"

//Define the setName() function.

void Player::setName(string par_name)

{

name = par_name;

}

//Define the setScore() function.

void Player::setScore(int par_score)

{

score = par_score;

}

//Define the getName() function.

string Player::getName()

{

return name;

}

//Define the getScore() function.

int Player::getScore()

{

return score;

}

7 0
3 years ago
You are stopped at a red traffic light and are first in line at the intersection. When the traffic light changes to green, you s
Ostrovityanka [42]

Answer:

Go, but only if the intersection is clear.

Explanation:

Traffic at intersection can be complicated at times. If the green light comes on after a red light, you have the right of way to go, but you should be careful to only go when the intersection is clear to avoid an accident. Once using the road, a good driver should be conscious of the other road users, as accidents might happen from you claiming your-right of-way

8 0
3 years ago
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
2 years ago
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