Answer:
3.6 kHz
Explanation:
The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.
The first harmonic will be as seen in the figure attached.
The length of the first harmonic will be λ/4.
λ/4=2.4 cm
λ=2.4 * 4=9.6 cm 0.096 m
Speed of Sound- 344 m/s(in air)
velocity(v) * Time Period(T) = Wavelength (λ)
Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}
\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency
Plugging in the values into the equation,
Frequency = Hz
= 3583.3 Hz≈3600 Hz= 3.6 kHz
Frequency= 3.6 kHz
The answer is:
71.6 <span>°F</span>
Answer:
the red at the bottom should not count, but the red at the top is the least dense because it floats upon the other liquids
Explanation:
hope this helps
Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).
Answer:
O a force that opposes motion