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I am Lyosha [343]
3 years ago
7

You watch a distant lady driving nails into her front porch at a regular rate of 1 stroke per second. You hear the sound of the

blows exactly synchronized with the blows you see. And then you hear one more blow after you see her stop hammering. How far away is she? The speed of sound is 340 m/s. a) 270 m. b) 680 m. c) 85 m. d) 1360 m. e) 170 m.
Physics
2 answers:
Evgen [1.6K]3 years ago
5 0

Answer:

The distance of the lady is 340 m

Solution:

As per the question:

Regular rate of stroke = 1 stroke per second

Speed of sound, v = 340 m/s

Now,

To calculate the distance of the lady, we need to calculate the wavelength of the sound wave which provides the distance.

Now, we know that:

v = \nu\times \lambda

where

\nu = frequency in Hz

\lambda = wavelength in meter

Also, we know that 1 stroke per second makes a frequency of 1 Hz

Thus

340 = 1\times \lambda

\lambda = 340\ m

Igoryamba3 years ago
4 0

Answer:

λ = 1360 m

Explanation:

Given data:

frequency of driving nails is given as 1 stroke per second mean at every 0.25 sec she hit the nails

speed of sound is given as 340 m/s

we know that the wave equation is given as

Speed = frequency × wavelength,

v = f × λ

where,

v = speed in meters/second (m/s)

f = frequency in Hertz (Hz)

substituing value to get wavelength  of her driving nails

340 m/s = (1Hz)\times  \lambda

\lambda = \frac{340}{0.25}

λ = 1360 m

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Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

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6 0
3 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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