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chubhunter [2.5K]
4 years ago
15

A stationary police car emits a sound of frequency 1200 hz that bounces off a car on the highway and returns with a frequency of

1250 hz. the police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) how fast was the moving car going?
Physics
1 answer:
Whitepunk [10]4 years ago
4 0

First, since the echo frequency is greater, the car must be roaming to the police car. 
This is for the reason that the car is "running over" its own wave fronts, creating the wavelengths tinier, and henceforth increasing the frequency. 

The trick here is to understand that the relative speed between the cars is half of the speed of the echo for the reason that the echo is returning to the police car at the similar speed as the sound wave trips to the moving car. 

It would also appear that the speed of sound has been expected to be 337 m/s 

So if v is the moving car's velocity, the correct formula to use is: 

2v / 337 = (1250/1200) - 1 
2v = 14.04 m/s 
v = 7.02 m/s

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Answer:

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Explanation:

Given the following data;

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Acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

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Substituting into the formula, we have;

Acceleration = \frac{15 - 24}{12}

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Acceleration = -0.75 m/s²

Therefore, the automobile is decelerating because its final velocity is lesser than its initial velocity, leading to a negative value.

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3 years ago
The production of carbon dioxide might be an indicator of what process
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How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
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Answer:

258774.9441 m

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y = Distance of probe from Sun

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G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

7 0
3 years ago
Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are se
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Here

\mu_0=  Permeability at free space constant

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d = Distance to the center point of the two object

Two magnetic field due to the current in the same directions then is,

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Replacing,

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B = 20*10^{-4}T

Therefore the correct answer is E.

8 0
3 years ago
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