Answer:
ΔX = λ = 0.68 m
Explanation:
Wave speed is related to wavelength and frequency by the equation
v = λ f
where the speed of sound is 340 m / s
λ = v / f
λ = 340/500
λ = 0.68 m
this is the wavelength, it is the minimum distance for which the wave epitates its movement, which is equal to the distance between two consecutive compressions of the sound
ΔX = λ = 0.68 m
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:
Explanation:
Expression for time period of a pendulum is as follows
T = 
l is length of pendulum from centre of bob and g is acceleration due to gravity
Given
Time period T = 1.583
g = 9.846
Substituting the values
1.583 = 
l = 
l = .6244 m
= 62.44 cm
Length of rod = length of pendulum - radius of bob
= 62.44 - 13.62
= 48.82 cm
= .488 m
Answer:
.
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is
(same as
,) meaning that
of energy is supplied to this lamp every second.
The question states that
out of that
of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the
of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive
of energy input and would outputs
of useful work. The efficiency of this lamp would be:
.