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2 years ago

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A fire engine is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the fire engin

e draws closer? The frequency increases, and the pitch decreases The frequency increases, and the pitch increases The frequency decreases, and the pitch decreases The frequency decreases, and the pitch increases

1 answer:

ValentinkaMS [17]2 years ago

7 0

Answer:

The frequency increases, and the pitch increases

Explanation:

Doppler's law of sound is applicable in such case when the observer or the sound source or both are moving relative to each other.

In such a case due to space-time constraint the waveform of the sound adjust themselves so as to obey the law of conservation of energy.

<u>The apparent frequency of the sound for the observer is given by:</u>

$f_o=(\frac{s+v_o}{s+v_s} )f$ ....................................(1)

where:

$f_o=$ observed frequency

$f=$ original source frequency

$s=$ speed of sound

$v_s=$ speed of source relative to the observer (taken negative when approaching towards the observer and vice-versa)

$v_o=$ speed of observer relative to the source (taken negative when moving away from the source and vice-versa)

<u>According to the given situation, eq. (1) becomes:</u>

$f_o=(\frac{s}{s-v_s} )f$

Since, $\frac{s}{s-v_s} >1$

Therefore

$f_o>f$

Pitch is very closely related to the frequency, it means that how fast is the amplitude of sound varying with time.

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Use the accompanying seismogram to answer which of the three types of seismic waves reached the seismograph first.

UkoKoshka [18]

Answer:

Primary waves (P-waves)

Explanation:

Due to excess of the energy inside the earth when the tectonic plates begin to slide or fracture then the energy is released in the form of seismic waves, this causes the earthquake.

<u>Two types of seismic waves are generally responsible for the earth quakes:</u>

body waves
surface waves

Body waves are of two types:

Primary waves (P-waves)

These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

These waves move along the surface of the earth. They are responsible for the earthquake's carnage.

They move up and down the Earth's surface, rocking the foundations of man-made structures.

Surface waves are slowest of the three waves, which means that they are the last to arrive. So at the end of an earthquake usually comes the most powerful shaking.

6 0

2 years ago

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

barxatty [35]

Answer:

a

The mass of blood is $m= 5.7876kg$

b

The number of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the question we are told that

The volume of blood is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of the blood is $\rho_b = 1060 kg/m^3$

% of blood that is cell is = 45.0%

% of the blood that is plasma is = 55.0%

density of blood cell is $\rho_d = 1125kg/m^3$

% of cell that are white is = 1%

% of cell that is red is = 99%

The diameter of the red blood cell is = $7.5 \mu m = 7.5*10^{-6}m$

The radius of the red blood cell is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

Generally the mass is mathematically represented as

$m = \rho_b * V_b$

Substituting value

$m = 1060 * 0.00546$

$m= 5.7876kg$

Mass of cell is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The volume of white blood cell is $V_w$ = 1% of volume of cells

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The volume of red blood cells is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The number of red blood cell is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The Number of white blood cell is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The total number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

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3 years ago

Read 2 more answers

How does the pupillary response prevent injury? What would happen without it?

Aleksandr [31]

Pupils dilate and constrict in order to allow an adequate amount of light to pass through the retina and vision. If there is not enough light and the pupils do not dilate, a small amount of light will pass to the retina and the vision will be damaged.

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3 years ago

Read 2 more answers

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst

disa [49]

Answer:

a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

v = u + at

v = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

v = u + at

v = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

s= ut + 0.5 at²

s = 0 x 1.8 + 0.5 x 42 x 1.8²

s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

s= ut + 0.5 at²

s = 0 x 3.6 + 0.5 x 42 x 3.6²

s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0

2 years ago