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3 years ago

How is the Doppler Effect used in astronomy and/or meteorology?

Physics

1 answer:

Maslowich3 years ago

4 0

Astronomers use the doppler effect to study the motion of objects across the Universe.

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3 years ago

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a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance

8_murik_8 [283]

The car's (average) acceleration would be

$a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}$

The car's position over time would be given by

$x=v_0t+\dfrac12at^2$

so that after 2.4 seconds, the car will have traveled a distance of

$x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2$

$\implies x=77.5\,\mathrm m$

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3 years ago

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A plant is thrown straight

olchik [2.2K]

The time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

<h3>How to calculate time?</h3>

The time taken for a motion to occur can be calculated using the following formula:

v² = u² - 2as

Where;

v = final velocity
u = initial velocity
s = distance
a = acceleration

8.84² = 0² + 2 × a × 7.01

78.15 = 14.02a

a = 5.57m/s²

V = u + at

8.84 = 0 + 5.57t

t = 1.59s

Therefore, the time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

Learn more about time at: brainly.com/question/13170991

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2 years ago

Which of the following is an organic compound?

AleksandrR [38]

The answer is D if im correct

8 0

4 years ago

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A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

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3 years ago