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} ] \\\\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C)
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 \\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.
Answer:
Kevin wants to make the firm private
Explanation:
Based on the information provided within the question in regards to the situation it seems that Kevin wants to make the firm private. Private Firms are companies that are owned by non-governmental entities or instead owned by a single individual or a very small amount of shareholders. Which is what Kevin seems to be wanting to do since he wants to buy back all the shares of the company so that only his family owns the company.
If you have any more questions feel free to ask away at Brainly.
There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.
What is the minimum runway length that will serve?
Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.
A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.
Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.
To learn more about runway
brainly.com/question/11553726
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Answer:
I do not have enough information to tell
Explanation:
This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.
Answer:
t= 137.5 s
Explanation:
So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another
<u>Slopes:</u>
Runner A: y = 7.50x + 55
Runner B: y = 7.90 x
sooooo
7.50 x + 55 = 7.90 x
- 7.50 x - 7.50 x
55 = .40 x
55/.40 = .40 x / .40
x = 137.5 s
t= 137.5 s
7.50 * 137.5 + 55 =1086.25 m
7.90 * 137.5 = 1086.25 m
