a train is running at 20m/s on a railway line with radius of curvature 40000 meters . the distance between the two rails is 15 meters . for safe running of train the elevation of outer rail over the inner rail is. (g= 10m/s)
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r is radius of the elevated path because distance between the two trains will acts as the radius of the banking of road.
r = 1.5
mv = 20 m/s
tanθ = v2rg where θ is the angle elevation.
tanθ = 20×201.5×10
= 26.66θ
= tan−(26.66)
r = 1.5
mv = 20 m/s
tanθ = v2rg where θ is the angle elevation.
tanθ = 20×201.5×10
= 26.66θ
= tan−(26.66)
adi131:
im sry but the answer is 1.50
Answered by
6
Answer: 1.50mm
Explanation:
Here use the formula
tan@=V^2/Rg.
V= velocity of train
R= radius of curvature
g= 10m/s^2
Therefore
tan@=20*20/40,000*10
tan@=1/1000
From triangle ABC
AB = elevation of outer rail
BC = distance between two rails
Therefore
tan@= AB/BC
1/ 1000 = AB/1.5
AB = 1.50*10^-3 m
AB = 1.50 mm
Therefore elevation of outer rail is 1.50mm.
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