Ask question

Ask question

All categories

1 year ago

15

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

its siren? The speed of sound on this day is 345 m/s.

1 answer:

Strike441 [17]1 year ago

8 0

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer $f_{obs}$ is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

$f_{obs} = f_s (\frac{v_w}{v_w-v_s})$

$f_s$ = Frequency of the source

$v_w$ = Speed of sound

$v_s$ = Speed of source

The velocity of the ambulance is

$v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 30.55m/s$

Replacing at the expression to frequency of observer we have,

$f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})$

$f_{obs} = 878Hz$

Therefore the frequency receive by observer is 878Hz

You might be interested in

Which are made of matter? Check all that apply. air bicycles sound waves people clothes sunlight trees

LUCKY_DIMON [66]

Trees bikes people clothes trees

5 0

1 year ago

Read 2 more answers

Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

earnstyle [38]

The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.

As the coefficient of absorption would define the energy present in the reflected wave, the material C has the highest percentage of absorption i.e. 62% and would be best suitable to make a sound proof room.

4 0

1 year ago

Read 2 more answers

A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work

SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string $T=5 N$

$2\pi r=2$

$r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m$

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is $90^{\circ}$

$W=\int\vec{F}\cdot \vec{r}$

$W=\int Fdr\cos 90$

$W=0$

4 0

1 year ago

Which statements identify what astronomers currently know and think will happen with our universe? Check all that apply. The big

Olegator [25]

The big bang produced dark energy, which accounts for some of the energy needed to expand the universe.

The vastness of space may contain a type of matter known as “dark matter.”

The universe is currently expanding at an accelerating rate.

Hope this helps !

7 0

1 year ago

Read 2 more answers

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

1 year ago