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The amplitude of the sine wave with RMS value of 220 V is
A = 220√2 volts.
The sine waveform is
v(t) = 220√2 sin(2πft)
where
f = 50 Hz, the frequency.
The period is
T = 1/f = 1/50 = 0.02 s
Use a graphical solution (shown below) to determine the number of times that v(t) = 220 in the interval t = [0, 0.02] s.
There are 2 instances when the voltage is 220 V in the interval t =[0, 0.02] s.
Note that 1 second is an integral multiple of 0.02 seconds.
Therefore in the interval [0,1], the number of instances when v(t) = 220 V is
(1/0.02)*2 = 100
Answer: 100
A = 220√2 volts.
The sine waveform is
v(t) = 220√2 sin(2πft)
where
f = 50 Hz, the frequency.
The period is
T = 1/f = 1/50 = 0.02 s
Use a graphical solution (shown below) to determine the number of times that v(t) = 220 in the interval t = [0, 0.02] s.
There are 2 instances when the voltage is 220 V in the interval t =[0, 0.02] s.
Note that 1 second is an integral multiple of 0.02 seconds.
Therefore in the interval [0,1], the number of instances when v(t) = 220 V is
(1/0.02)*2 = 100
Answer: 100
Answer:
Detailed solution is given below:
Answer:
972π in³
Step-by-step explanation:
The volume (V) of a sphere is calculated as
V = πr³ ← r is the radius
Here r = 9, thus
V = π × 9³ =
π × 729 ( divide 3 and 729 by 3 )
= 4π × 243
= 972π in³