Answer:
Explanation:
Given that
Vrms= 60 V
Vf= 0.7 V
We know that peak value of AC voltage given as
Now by putting the values
The output voltage of the DC current given as
Therefore output voltage of the DC current is 84.15 V.
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
1 answer:
Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
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Answer:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Explanation:
A cubic has the form:
y = ax³ + bx² + cx + d
Given four points, we can write a system of equations:
1 = a + b + c + d
1/2 = 8a + 4b + 2c + d
1/3 = 27a + 9b + 3c + d
1/4 = 64a + 16b + 4c + d
Solving this algebraically would be time-consuming, but we can use matrices to make it easy.
First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:
Now we multiply by the solution matrix (again using a calculator):
So the cubic is:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12