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AleksandrR [38]

4 years ago

13

A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 37.5

kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.021 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

1 answer:

n200080 [17]4 years ago

5 0

Answer:

39.11 Hz

Explanation:

Data provided in the question:

Frequency of sound emission, $f_0$ = 37.5 kHz = 37500 Hz

Speed of bat, $v_{bat}$ = 0.021 times the speed of sound

Now,

Frequency heard by bat = $f_0\times(\frac{v+v_{bat}}{v-v_{bat}})$

Therefore,

The Frequency heard by bat will be = $37500\times(\frac{343+0.021\times343}{343-0.021\times343})$

or

Frequency heard by bat will be= $37500\times\frac{350.203}{335.797}$

or

Frequency heard by bat will be = 39108.78 Hz = 39.11 Hz

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A plug-in transformer supplies 9 V to a video game system. (a) How many turns are in its secondary coil, if its input voltage

zloy xaker [14]

Answer:

The number of turns in the secondary coil is 28.

Explanation:

Given that,

Input voltage given to the transformer, $V_p=111\ V$

Output voltage supplied to the video game, $V_s=9\ V$

Number of turns in the primary coil, $N_p=340$

We need to find the number of turns in its secondary coil. We know that the transformer equation is given by :

$\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}\\\\N_s=\dfrac{V_s}{V_p}\times N_p\\\\N_s=\dfrac{9}{111}\times 340\\\\N_s=27.56$

or

$N_s=28$

So, the number of turns in the secondary coil is 28.

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3 years ago

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it is a property of motion

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3 years ago

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3 years ago

A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it

Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

$I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}$

The required initial angular speed is obtained herein:

$g= \omega_{o}^{2}\cdot R_{ss}$

$\omega_{o}=\sqrt{\frac{g}{R_{ss}} }$

$\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }$

$\omega_{o} \approx 0.253\,\frac{rad}{s}$

The initial moment of inertia is:

$I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}$

The final moment of inertia is:

$I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}$

Now, the final angular speed is obtained:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )$

$\omega_{f} = 0.331\,\frac{rad}{s^}$

The apparent acceleration is:

$a_{f} = \omega_{f}^{2}\cdot R_{ss}$

$a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)$

$a_{f} = 16.763\,\frac{m}{s^{2}}$

This is approximately 1.709g.

4 0

4 years ago

The speed of a transverse wave on a string is 450 m/s, and the wavelength is 0.19 m. The amplitude of the wave is 1.6 mm. How mu

tigry1 [53]

Answer: 2.22s

Explanation: wave speed = 450 m/s, A = amplitude = 1.6mm, λ= wavelength = 0.19m

Wave speed = distance traveled / time taken

Distance traveled = 1km = 1000 m

450 = 1000/ t

t = 1000/ 450 = 2.22s

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3 years ago