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Brut [27]
3 years ago
10

A galloping pony speeds past you at 5 m/s. The frequency of the sound produced by the hooves on the dirt is 221 Hz. Assume the s

peed of sound to be 342 m/s. What is the observed frequency of the hooves after the pony has passed your position?
1. 218 Hz
2. 226 Hz
3. 216 Hz
4. 224 Hz
Physics
2 answers:
kow [346]3 years ago
7 0
Given:
speed of passing pony 5 m/s
frequency of the sound produced: 221 Hz
speed of sound 342 m/s

Let us use the Doppler Shift Formula:
Where the <span>source is moving away from the observer at rest
</span>
f' = (v / v+vs) f
Where, vs<span> = Velocity of the Source,</span>
           v = Velocity of sound or light in medium,
           f = Real frequency,
           f' = Apparent frequency.

f'= [342 m/s / (342 m/s+5m/s)] * 221 Hz
f' = 0.9856 * 221Hz
f' = 217.8176 Hz or 218 Hz

The observed frequency <span>of the hooves after the pony has passed your position is 218 Hz.</span>
Studentka2010 [4]3 years ago
7 0
The answer is A) 218 Hz
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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
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Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

R=75 cm=\frac{75}{100}=0.75 m

1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

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The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g
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Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

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a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

4 0
3 years ago
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti
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Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

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distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

d = v. t + \dfrac{v^2}{2a}

t = \dfrac{d}{v} -\dfrac{v}{2a}

t is the reaction time

t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}

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hence, the reaction time of the ranger is equal to 1.58 s.

3 0
3 years ago
Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc
Juli2301 [7.4K]

Answer:

0.2286  m,  0.686 m and  1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

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where r is the distance,  λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

        v =  λ f

         λ = v / f

we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

     Δr = (2n + 1) 0.2286

we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

     2    1,143

therefore we see that there is respect even where the intensity is minimal

6 0
3 years ago
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