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
Answer:
3.6 kHz
Explanation:
The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.
The length of the first harmonic of a closed pipe is given as;
L = λ/4 -------------------- (i)
where L = Length
and λ = wavelength
2.4 = λ/4
λ = 2.4 x 4 = 9.6 cm
Also, v = fλ ------------------ (ii)
where v = speed of sound in air = 344 m/s
f = frequency of wave in Hertz
f = v/λ ------------------ (iii)
convert 9.6 cm to m = 0.096 m
substitute for λ and v in (iii)
3583.33 Hz = 3600 Hz = 3.6 kHz
Answer:
Acceleration due to gravity will be
Explanation:
We have given length of pendulum l = 55 cm = 0.55 m
It is given that pendulum completed 100 swings in 145 sec
So time taken by pendulum for 1 swing
We have to find the acceleration due to gravity at that point
We know that time period of pendulum;um is given by
So
Squaring both side
So acceleration due to gravity will be
Answer:
Magnitude of angular acceleration = -3.95 rad/s²
Explanation:
Angular acceleration is the ratio of linear acceleration and radius.
That is
Radius = 72 cm = 0.72 m
Linear acceleration is rate of change of velocity.
Angular acceleration
Angular acceleration = -3.95 rad/s²
Magnitude = 3.95 rad/s²
Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁ + m₂
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁ + m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂
) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly