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MAXImum [283]
3 years ago
13

The drag coefficient of a vehicle increases when its windows are rolled down or its sunroof is opened. A sports car has a fronta

l area of 18 ft2 and a drag coefficient of 0.32 when the windows and sunroof are closed. The drag coefficient increases to 0.41 when the sunroof is open. Determine the additional power consumption of the car when the sunroof is opened at (a) 35 mi/h and (b) 70 mi/h. Take the density of air to be 0.075 lbm/ft3 .

Engineering
2 answers:
Ne4ueva [31]3 years ago
5 0

Answer:

the additional power consumption is a) 0.4669 hp, b) 3.7353 hp

Explanation:

the solution is in the attached Word document

Download docx
klio [65]3 years ago
3 0

Answer:

Explanation:

The additional power consumption of the car when v :

35 mi/h = 0.464hp

70 mi/h = 3.71hp.

For the step by step explanation of how we arrived at these answers, please go through the attached files.

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how to calculate the torque when a force is applied on a cog? explain the step-by-step and provide an illustration/diagram. Can
yawa3891 [41]

Explanation:

Torque is the cross product of the radius vector and force vector:

τ = r × F

In other words, it is equal to the radius times the perpendicular component of the force.

τ = r · Ftangential

If we call θ the angle between the radius and the force, then:

τ = r · F sin θ

3 0
4 years ago
A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is
Nataly [62]

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet  = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

5 0
4 years ago
Subcooled liquid water flows adiabatically in a constant diameter pipe past a throttling valve that is partially open. The liqui
Llana [10]

Answer:

hi-he = 0

pi-pe  = positive

ui-ue = negative

ti-te = negative

Explanation:

we know that fir the sub cool liquid water is

dQ = Tds = du +  pdv   ............1

and  Tds = dh - v dP         .............2

so now for process of throhling is irreversible when v is constant

then heat transfer is = 0 in irreversible process

so ds > 0

so here by equation 1 we can say

ds  > 0  

dv = 0 as v is constant

so that Tds = du    .................3

and du > 0

ue - ui > 0

and

now by the equation 2 throttling process  

here enthalpy is constant

so dh = 0

and Tds  = -vdP

so ds > 0  

so that -vdP > 0  

as here v is constant

so -dP =P1- P2

so P1-P2 > 0

so pressure is decrease here

5 0
3 years ago
Before finishing and installing a shelved cabinet you just constructed, you need to check the
Greeley [361]

Answer:

Carpenter's square

Explanation:

The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.

Other hand tools that are used to measure angles are;

  • The combination square that allows a user to set both 90°  and 45° angles
  • A Bevel that allows users to set any angle they like.
  • A Protractor that resembles a bevel but its marks are marked in an arc.
  • An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.
7 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

8 0
3 years ago
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