Answer:
a. Heat removal rate will increase
b. Heat removal rate will decrease
Explanation:
Given that
One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.
We know that heat transfer in fin given as follows
![Q_{fin}=\sqrt{hPKA}\ \Delta T](https://tex.z-dn.net/?f=Q_%7Bfin%7D%3D%5Csqrt%7BhPKA%7D%5C%20%5CDelta%20T)
We know that area
![A=\dfrac{\pi}{4}d^2](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cpi%7D%7B4%7Dd%5E2)
Now when diameter will triples then :
![A_f=\dfrac{\pi}{4}{\left (3d \right )}^2](https://tex.z-dn.net/?f=A_f%3D%5Cdfrac%7B%5Cpi%7D%7B4%7D%7B%5Cleft%20%283d%20%5Cright%20%29%7D%5E2)
![A_f=9A](https://tex.z-dn.net/?f=A_f%3D9A)
![Q'_{fin}=\sqrt{9hPKA}\ \Delta T](https://tex.z-dn.net/?f=Q%27_%7Bfin%7D%3D%5Csqrt%7B9hPKA%7D%5C%20%5CDelta%20T)
![Q'_{fin}=3\sqrt{hPKA}\ \Delta T](https://tex.z-dn.net/?f=Q%27_%7Bfin%7D%3D3%5Csqrt%7BhPKA%7D%5C%20%5CDelta%20T)
![Q'_{fin}=3Q](https://tex.z-dn.net/?f=Q%27_%7Bfin%7D%3D3Q)
So the new heat transfer will increase by 3 times.
Now when copper rod will replace by aluminium rod :
As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.
Answer:
A sole proprietorship, also known as the sole trader, individual entrepreneurship or proprietorship, is a type of enterprise that is owned and run by one person and in which there is no legal distinction between the owner and the business entity.
Explanation:
Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
![E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\](https://tex.z-dn.net/?f=E%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%20%2B6.32%2B6.33%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20A%20%29%20%3D%206.34%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%20%2B6.31%2B6.32%2B6.35%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%206.3225%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%20%2B6.32%2B6.36%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%206.355%5C%5C)
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
![Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\](https://tex.z-dn.net/?f=Var%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%5E2%2B6.32%5E2%2B6.33%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.34%5E2%5C%5C%5C%5CVar%20%28%20A%20%29%20%3D%200.00025%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%5E2%2B6.31%5E2%2B6.32%5E2%2B6.35%5E2%7D%7B4%7D%20-%206.3225%5E2%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%200.00026875%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%5E2%2B6.32%5E2%2B6.36%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.355%5E2%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%200.000475%5C%5C)
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
![P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\](https://tex.z-dn.net/?f=P%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.34-6.321%7D%7B6.321%7D%2A100%3D%200.30058%5C%5C%5C%5CP%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.3225-6.321%7D%7B6.321%7D%2A100%3D%200.02373%5C%5C%5C%5CP%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.355-6.321%7D%7B6.321%7D%2A100%3D%200.53788%5C%5C)
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate
Answer:
178 kJ
Explanation:
Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:
ΔU = -Q = -c*m*ΔT (1)
where c= specific heat of water = 4180 J/kg*ºC
m= total mass = 6*0.355 Kg = 2.13 kg
ΔT = difference between final and initial temperatures = 20ºC
Replacing by these values in (1), we can solve for Q as follows:
Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ
So, the amount of heat transfer from the six canned drinks is 178 kJ.
Answer:
Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z
Explanation:
The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.
A detailed step by step explanation is attached below.