Select the best answer for the question.

Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has

trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0

2 years ago

Use the Hurricanes data and via Multiple regression select the three input variables: Min-pressure, Gender, Category in order to

S_A_V [24]

Answer:

Answer for the question:

Use the Hurricanes data and via Multiple regression select the three input variables: Min-pressure, Gender, Category in order to predict the All-Deaths. Pick one of the three variables that has the best P-value and re-do the regression using ONLY this one variable to predict the All-deaths. Answer the questions. (Numerical answers are rounded so choose the answer that matches the best).

is explained in the attachment.

Explanation:

Download pdf

6 0

3 years ago

Question 2) A material that is malleable can be defined as:

Reil [10]

Answer:

A.

Explanation:

Able to return to it's original shape after distortion.

7 0

2 years ago

Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.1

Tamiku [17]

Answer:

(a) the mass flow rate of air is 5.351 kg/s

(b) the input power required is 2090.786 kW

Explanation:

Given;

initial pressure, P₁ = 100 kPa

initial temperature, T₁ = 20 °C

Final pressure, P₂ = 1.8 MPa

Final temperature, T₂ = 400 °C

Inlet area of the compressor = 0.15 m²

outlet area of compressor = 0.078 m²

velocity of air = 30 m/s

Part (a) mass flow rate of air through the inlet

Mass flow rate = Area x velocity = density x volumetric rate

m = Av = ρV

from ideal gas law, PV = nRT and ρ = m/V

substitute these values in the above equations, we will have;

$m = \frac{PAv}{RT}$

$m = \frac{100000*0.15*30}{287*(20+273)}= 5.351 \ kg/s$

Part (b) the required power input

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2$

where;

W is the input power

m is the mass flow rate

h₁ is the initial enthalpy

h₂ is the final enthalpy

initial and final enthalpy are obtained from steam table using interpolation;

h₁ = 293.166 kJ

h₂ = 684.344 kJ

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2\\\\W = mh_2 - m(h_1+\frac{v_1{^2}}{2})\\\\W = 5.351 ( 684.344) - 5.351 (293.166 + \frac{30^2}{2000}) \\\\W = 3661.925 \ kW -1571.139 \ kW\\\\W = 2090.786 \ kW$

7 0

2 years ago

Read 2 more answers

For java

fgiga [73]

Answer:

Explanation:

Hi,

I have updated the code. Hghlighted the code changes below

CourseGradePrinter.java

import java.util.Scanner;

public class CourseGradePrinter {

public static void main (String [] args) {

final int NUM_VALS = 4;

int[] courseGrades = new int[NUM_VALS];

int i = 0;

courseGrades[0] = 7;

courseGrades[1] = 9;

courseGrades[2] = 11;

courseGrades[3] = 10;

for(i=0; i<NUM_VALS; i++){

System.out.print(courseGrades[i]+" ");

}

System.out.println();

for(i=NUM_VALS-1; i>=0; i--){

System.out.print(courseGrades[i]+" ");

}

return;

}

Output:

7 9 11 10

10 11 9 7

5 0

2 years ago