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1 answer:
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Answer:
a) 30.82 m/s
b) 183.33 Hz
Explanation:
a)
V = speed of the sound = 339 m/s
v = speed of the train = ?
f' = observed frequency by the observer = 220 Hz
f = actual frequency of the observer = 200 Hz
using the equation
v = 30.82 m/s
b)
V = speed of the sound = 339 m/s
v = speed of the train = 30.82 m/s
f'' = observed frequency by the observer as train moves away = ?
f = actual frequency of the observer = 200 Hz
using the equation
f'' = 183.33 Hz
Answer:
the unloaded braking efficiency is 84.6 %
Explanation:
Given the data in the question;
by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula
S = (Y( V₁² - V₂²)) / ( 2g( ηbμ +
± sin∅
))
now given that the tracked is levelled, ∅ = 0, also Y
= 1.04 for level or flat road
Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s
now, we substitute in our values to get the braking efficiency;
180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)
180ft = 8053.76 / ( 0.5152ηb + 1.1592)
180( 0.5152ηb + 1.1592) = 8053.76
( 0.5152ηb + 1.1592) = 8053.76 /180
0.5152ηb + 1.1592 = 44.7431
0.5152ηb = 44.7431 - 1.1592
0.5152ηb = 43.5839
ηb = 43.5839 / 0.5152
ηb = 84.596 ≈ 84.6 %
Therefore, the unloaded braking efficiency is 84.6 %