What can we learn from space images?

The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to t

kompoz [17]

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 = $\frac{344}{0.096}$ Hz

= 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

7 0

2 years ago

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Consider a system of a cliff diver and the earth. the gravitational potential energy of the system decreases by 28,000 j as the

olga2289 [7]

The decrease in gravitational potential energy of the system is given by

$\Delta U = (mg) \Delta h$

where

m is the mass, g is the gravitational acceleration, and $\Delta h$ is the variation of height of the system.

$(mg)$ also corresponds to the weight of the diver, therefore if we rearrange the equation and we use $\Delta U=28000 J$ and $\Delta h=32.0 m$ , we can find her weight:

$mg=\frac{\Delta U}{\Delta h}=\frac{28000 J}{32 m}=875 N$

5 0

3 years ago

The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end

Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

$T = mg$

here we will have

$T = 2(9.8) = 19.6 N$

now we will have speed of wave given as

$v = \sqrt{\frac{T}{\mu}}$

here we will have

$v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}$

$v = 16.33 m/s$

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

$\lambda = \frac{v}{F}$

$\lambda = \frac{16.33}{100} = 0.16 m$

so wavelength will be 0.16 m

5 0

3 years ago

When you catch a baseball, what kind of work do you do, negative or positive? By catching the baseball, what do you change? Does

bagirrra123 [75]

Negative energy by catching it. Changes the force and movement of the baseball. Loses energy. Kinetic energy

3 0

2 years ago

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A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago