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ANTONII [103]
3 years ago
14

HOLA COMO ESTAN TODOS

Engineering
1 answer:
ella [17]3 years ago
4 0

Answer:

buenos diaz, coma estas?

Explanation:

You might be interested in
The primary heat transfer mechanism that quickly warms my hand if I hold it directly above a campfire is: a)-Radiation b)-Induct
Tanzania [10]

Answer:

The correct answer is option 'c':Convection.

Explanation:

When we ignite a campfire the heat produced by combustion heats the air above the fire. As we know that if a gases gains heat it expands thus it's density decreases and hence it rises, if we hold our hands directly above the fire this rising hot air comes in contact with our hands thus warming them.

The situation is different if we are at some distance from the campfire laterally. Since the rising air cannot move laterally the only means the heat of the fire reaches our body is radiation.

But in the given situation the correct answer is convection.  

6 0
3 years ago
Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate
Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy +  flow work

E = \dot m ( \frac{v^2}{2] +  zg + p\nu)

E = \dot m( \frac{v^2}{2] +  zg + p_{atm} \frac{1}{\rho})

57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})

solving for flow rate

\dot m = 99.977we know that [tex]\dot m  = \rho AV

\dot m  = 780 \frac{\pi}{4} D^2\times 16

solving for d

99.97 = 780 \times \frac{\pi}{4} D^2\times 16

d = 0.090 m

so radius = 0.045 m

3 0
4 years ago
1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus
Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

7 0
3 years ago
thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat
eduard

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

7 0
3 years ago
What scale model proves the initial concept?
Tju [1.3M]

Answer: A prototype

Explanation:

5 0
3 years ago
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