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ladessa [460]
2 years ago
8

Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee

ding car, Officer O'Malley jumps in his police car and turns on the siren, whose frequency is 800. Hz. If the officer chases after the Gonzalez' car with a speed of 35.0 m/s, what frequency do the Gonzalezes hear as the officer approaches?

Physics
1 answer:
valina [46]2 years ago
3 0

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

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A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki
Anna71 [15]
Based on the direction of propagation compared to direction of vibration, waves are classified into:
1- Transverse waves: The direction of propagation of the wave is perpendicular to the direction of vibration of the medium particles.
2- Longitudinal waves: The direction of propagation of the wave is the same as the direction of vibration of the medium particles.

For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave
6 0
2 years ago
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
2 years ago
Read 2 more answers
A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkin
ra1l [238]

Answer: The power is 156 watt

Explanation:

is in the attachment

3 0
2 years ago
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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova
Mrac [35]

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
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