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Virty [35]
4 years ago
5

A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H

z) is heard by a boy on a bicycle riding away from the car at 8.25 m/s? Take the speed of sound to be 343 m/s.
Physics
1 answer:
IRINA_888 [86]4 years ago
8 0

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

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The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the spe
Snezhnost [94]

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

5 0
3 years ago
A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertical
Sonbull [250]

Answer:

a. t=2.997\ s is the time after which the two stone hit the water surface.

b. u_2=1.998\ m.s^{-1}

c. v_1=31.3688\ m.s^{-1}   &  v_2=31.3686\ m.s^{-1}

Explanation:

Given:

  • height of the cliff, h=50\ m
  • time gap between the projection of stones, \Delta t=1\ s
  • initial velocity of first stone, u_1=+2\ m.s^{-1}

Here positive sign means that the stone is thrown vertically downward in the direction of gravity.

a.

Using equation of motion:

h=u_1.t+\frac{1}{2} g.t^2

50=2t+4.9t^2

t=2.997\ s is the time after which the two stone hit the water surface.

b.

using the equation of motion for second stone:

h=u_2.t+\frac{1}{2} g.t^2

50=u_2\times 2.997+4.9\times 2.997^2

u_2=1.998\ m.s^{-1}

c.

Final velocity of the stone at the instant of touching the water surface:

v_1^2=u_1^2+2.g.h

v_1^2=2^2+2\times 9.8\times50

v_1=31.3688\ m.s^{-1}

&

v_2^2=u_2^2+2g.h

v_2^2=1.998^2+2\times 9.8\times 50

v_2=31.3686\ m.s^{-1}

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