The power in a lightbulb is given by the equation P- FR where /is the current flowing through the lightbulb and Ris the resistan
2 answers:
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Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = =
= 18.38 rad/sec
Period of oscillation =
= 0.34 sec
Answer:
Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.
Explanation:
We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.
<u>According to the Ohm's law, we find its resistance:</u>
So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.
Here the value of frequency is the number of times the current changes its direction or the polarity in one second.