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valina [46]
2 years ago
13

Consider coaxial, parallel, black disks separated a distance of 0.20 m. The lower disk of diameter 0.40 m is maintained at 500 K

and the surroundings are at 300 K. What temperature will the upper disk of diameter 0.20 m achieve if electrical power of 17.5 W is supplied to the heater on the back side of the disk?
Engineering
1 answer:
ANTONII [103]2 years ago
4 0

Answer:

In this picture is the answer

Explanation:

You might be interested in
How many robots does bailey nursery own ​
givi [52]

Answer:

The Bailey family has flourished during its business’ 110-year history. But Bailey Nurseries’ leaders still operate with the belief that the family doesn’t always know best. The company has grown from a one-man operation selling fruit trees and ornamental shrubs to one of the largest wholesale nurseries in the United States, thanks to insights from those who are family and those who aren’t.

“For a business to thrive, you have to ask for outside help,” says Terri McEnaney, president of the Newport-based company and a fourth-generation family member. “We get an outside perspective through family business programs, advisors and our board, because you can get a bit ingrained in your own way of thinking.”

When Bailey Nurseries chose its current leader in 2000, it brought in a facilitator who gathered insights from key employees, board members and owners. Third-generation leaders (and brothers) Gordie and Rod Bailey picked Rod’s daughter McEnaney, who had experience both inside and outside the company.

Explanation:

5 0
2 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
Sawing stock to reduce its thickness is known as __________ .
rjkz [21]

Answer:

resawing

Explanation:

8 0
3 years ago
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
The way most recursive functions are written, they seem to be circular at first glance, defining the solution of a problem in te
EastWind [94]

Question Continuation

int factorial(int n) {

if(n == 0)

return 1;

else

return n * factorial(n - 1);

}

Provide a brief explanation why this recursive function works.

Show all steps involved in calculating factorial(3) using the function defined.

Answer:

1. Brief explanation why this recursive function works.

First, the recursive method factorial is defined.

This is the means through with the machine identifies the method.

The method is defined as integer, the machine will regard it as integer.

When the factorial is called from anywhere that has access to it, which in this case is within the factorial class itself. This means you can call it from the main method, or you can call it from the factorial method itself. It's just a function call that, well, happens to call itself.

2. Steps to calculate factorial(3)

1 First, 3 is assigned to n.

2. At line 2, the machine checks if n equals 0

3. If yes, the machine prints 1

4. Else; it does the following from bottom to top

factorial(3):

return 3*factorial(2);

return 2*factorial(1):

return 1;

Which gives 3 * 2 * 1 = 6

5. Then it prints 6, which is the result of 3!

6 0
3 years ago
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