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hammer [34]
1 year ago
8

4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember

, you can assume the following things: The whistle you use to call your hunting dog has a frequency of 21.0 kHz, but your dog is ignoring it. You suspect the whistle may not be working, but you can't hear sounds above 20.0 kHz. The speed of sound is 330 m/s at the current air temperature.
Physics
1 answer:
jeyben [28]1 year ago
8 0

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

Explanation:

The Doppler shift equation is given as follows;

f' = \dfrac{v - v_o}{v + v_s} \times f

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

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A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is
kirill115 [55]

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

\mu = Coefficient of friction

\theta = Slope

From Newton's second law

mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655

Applying \mu to the above equation and \theta=10^{\circ}

mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

3 0
1 year ago
The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fractio
Alexxandr [17]

Answer:

4/10 of L.

Explanation:

A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.

The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.

The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)

For an open pipe,

Fundamental  mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5×2l₀ = 4L× 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i.e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10(L)

∴ 4/10 of L must be cut off

7 0
2 years ago
Which best describes what forms in nuclear fission?A. two smaller, more stable nucleiB. two larger, less stable nucleiC. one sma
Step2247 [10]

Answer:B. two larger, less stable nuclei

Explanation: They collied and don't combine

4 0
2 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
2 years ago
How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?
AnnZ [28]
Molecules in a gas move faster than in a liquid.

hope it helps.
3 0
1 year ago
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