Answer:
Explanation:
Given that,
AC frequency of 2.3KHz
f=2.3×10³Hz
Vrms produce is
Vrms=1.5V
Current rms
Irms= 31mA
The capacitor is reconnected to a generator of frequency
f=4.8KHz =4800Hz
The current rms becomes
Irms= 85mA
Vrms=?
Solution
First genrator
The capacitive reactance is given as
Xc=Vrms/Irms
Xc=1.5/31×10^-3
Xc=48.39 ohms
Now, to know the capacitance of the capacitor
Xc=1/2πfC
Then,
C=1/2πfXc
So,
C=1/2×π×2300×48.39
C=1.43×10^-6C
C=1.43μF
Note: the capacitance of the capacitor did not change,
Now for generator two.
The reactance are given as
Xc=1/2πfC
Xc=1/2×π×4800×1.43×10^-6
Xc=23.19ohms
Then,
Vrms2=Irms2 ×Xc
Vrms2=85×10^-3×23.19ohms
Vrms2=1.97V
Vrms2=1.97Volts
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Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula

Option (A) is correct.
As we know that power is defined as rate of work done
so we will have

so in order to increase the power as per above formula we know that either we need to increase the work or we need to decrease the time to complete that work
So here the correct answer will be
increase the work being done or decrease the time in which the work is completed.