Answer:
1) 63.66 ohm
2) 188.49 ohm
Explanation:
Data provided in the question:
Part 1
Capacitance, C = 5μF = 5 × 10⁻⁶ F
Frequency = 500 Hz
Now,
Impedance = 
or
Impedance = 
or
Impedance = 63.66 ohm
Part 2
Inductance = 60 mH = 60 × 10⁻³ H
Frequency = 500 Hz
Now,
Impedance for an inductor = 2πfL
thus,
Impedance = 2 × π × 500 × 60 × 10⁻³
= 188.49 ohm
Answer:
Area required = 9.5 ft²
Explanation:
Step by step explanation is given in the attached document.
Answer:
Total saving would be of 36.917 $\yr
Explanation:
Given Data:
Degree F
Degree F
A = 30 ft^2
%
Heat loss before previous final value 

= 360 Btu/hr
Heat loss after new value
saving would be 
= 616.1782 kw hr/yr
$
= 36.917 $\yr
Answer:
I think D is correct
Explanation:
C is decreasing function, probably worst
A is arctan -> in radian, the rate of increasing is very slow-> second worst
B(14) = ln(9*14) = 4.8
D(14) = sqrt(8+14^2)=14.2
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