The real peak voltage is 120/0.707 = 170 V.
The peak voltage is the highest point or voltage value in any voltage waveform. A power quality issue arises when Pulse Width Modulation (PWM) devices, such as variable frequency drives, are added to a power system with a peak voltage equal to the square root of two times the RMS voltage. The peak voltage, for instance, if the RMS voltage is 85 V. The average voltage and maximum voltage of AC power coming from the wall are both about 110 V. Therefore, the real peak voltage is 120/0.707 = 170 V. The sinusoid's amplitude is divided in half by this. Peak-to-peak voltage (Vp-p), also known as the total amplitude, is 340 V, or twice the peak voltage.
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In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.
Explanation:
Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.
Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.
By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.
After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.
Answer:
The answer for this solution is 4
Explanation:
MA=L/E
800/200
=4
Answer:
<h2>
2113 seconds</h2>
Explanation:
The general decay equation is given as
, then;
where;
is the fraction of the radioactive substance present = 1/16
is the decay constant
t is the time taken for decay to occur = 8,450s
Before we can find the half life of the material, we need to get the decay constant first.
Substituting the given values into the formula above, we will have;

Half life f the material is expressed as 


Hence, the half life of the material is approximately 2113 seconds