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3 years 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]3 years 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

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Radio stations have dipole type antennas

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4 years ago

How do you find average velocity

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Answer:

find the sum of the inital and final velocitys and divide by 2 to find the average

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3 years ago

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh

SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

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170.097 = 46451.5 x t x 2.70

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3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

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3 years ago

Read 2 more answers

A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a

MAXImum [283]

Answer:21.18 m

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Given

initial speed u=10 m/s

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time taken to complete 22 m

$h=ut+\frac{1}{2}at^2$

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$22=\frac{1}{2}gt^2$

$t=\sqrt{\frac{22\times 2}{g}}$

$t=2.11 s$

Horizontal Distance moved

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$R=10\times 2.11$

$R=21.18 m$

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3 years ago