As the source approaches you, the sound waves are compressed, so
the pitch of the sound is higher than what the source is actually emitting.
Then, after it passes you and begins moving away, the sound waves
are stretched, so the pitch of the sound is lower than what the source
is actually emitting.
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
No waves because Q19 waves would going at the surface at regions
Answer:
The inventor of the electric cell was:
Alessandro Volta (in other words, Volta)
Explanation: