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2 years ago

11 Notează, în caiet, trăsăturile personajelor ce se pot

1 answer:

6 0

Answer:a.Bianca:”Puiul de Urs loveste cu labele din fata in roata de cauciuc atarntata de-o sfoara”:Raspuns:Trasaturile puiului de urs sunt:jucaus,curios,interesat

b.Ingrijitoarea:”Doar un copil,asta esti matalută.Ia la mama!Stiu,usor nu ti o fi nici tie....”Raspuns:Trasaturile ingrijitoarei sunt: Iubitoare de animale,buna la suflet,adica ea este foarte atenta,impresionata de puiul de urs si de animale.

c.Mara:”Of,mama o considera si acum tot o fetita,desi a terminat facultatea in vara asta.”Raspuns:Trasaturile Marei sunt:Neintelegatoare,incapatanata,adica nu asculta de mama ei,crede ca dansa o considera tot o fetita,si fata nu este de fapt asa(ti am dat un exemplu.)

Explanation:

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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:

Error.java

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

2 years ago

To water his lawn, a homeowner uses two hoses. One connects to the faucet, the other to the end of the first hose to make the ho

Shtirlitz [24]

Answer: to be exact you need 28mm of tubing for that

Explanation:

When the election

8 0

2 years ago

Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige

Lera25 [3.4K]

Answer:

A) $COP = \frac{16.97}{9.8} = 1.731$

B) $P_{IN} = 0.4763$

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

$T_1 = 28$ degree celcius

$T_2 = 12$ degree celcius

$\dot m = 1140 kg/h$

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

$A) Q = \dot m Cp \Delta T$

$= 1140 \times 3.35\times (28-12) = 61,104 kJ/h$

$Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec$

$COP = \frac{16.97}{9.8} = 1.731$

$COP ∝ \frac{1}{P_{in}}$

$P_{in}$ wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

$COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625$

we know that

$COP = \frac{heat\ obsorbed}{P_{in}}$

$P_{IN} = \frac{16.97}{35.62} = 0.4763$

c) second law efficiency

$\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%$

exergy destruction os given as $X = W_{IN} - X_{Q2}$

= 9.8 - 0.473 = 9.3237 kW

8 0

2 years ago

print('Predictions are hard.") print(Especially about the future.) user_num = 5 print('user_num is:' user_num)

Bas_tet [7]

Answer:

The correct code is given below:-

print("Predictions are hard.")

print("Especially about the future.")

user_num = 5

print("user_num is:", user_num)

7 0

2 years ago

A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 1.5 ft3/s

svetoff [14.1K]

Answer:

11.52 hp

Explanation:

Givens:

p_1 = 15 pisa

p_2 = 70 pisa

V_ol=1.5 ft^3/s

Solution:

Note: m = p x V_ol (assuming in compressible flow —> p =const)

The total change in the system mechanical energy can be calculated as follows,

Δ e= (p_2 - p_1 ) /p

The power needed can be calculated as follows

P = W =mΔ e = p x V_ol x(p_2 - p_1 ) /p

= V_ol x (p_2 - p_1 )

= 44 pisa. ft^3/s

= 44 x (1 btu/5.404pisa. ft^3) x (1 hp/0.7068btu/s)

= 11.52 hp

3 0

2 years ago