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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.
Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it
Answer:
R = 710.7N
L = 67.689 N
During gravity fall L = R = 0 N
Explanation:
So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is
g + 0.15g = 1.15g
The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):
a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N
The force L that she exerts on the package would be:
am = 1.15g *6 = 1.15*9.81*6 = 67.689N
When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.
The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.
The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.
<h3>What is a Circuit ?</h3>
In electronics, a circuit is a complete circular conduit through which electricity flows. A simple circuit consists of conductors, a load, and a current source. The term "circuit" broadly refers to any continuous path via which electricity, data, or a signal might flow.
- The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.
Hence, flow control circuit design that can best control an overrunning load is the opposing circuit
Learn more about Circuit here:
brainly.com/question/26064065
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