find the angle of banking of curved railway track of radius 600m if the maximum saftey sppedd limit is 54;km/hr .if distance beetween the rails is 1.6m find elevation ofvthe outervtrack above the inner track

A railway? Don't know about a railway. If we're talking about using the normal force to keep circular motion we're talking about sin-1(v^2/(gr)) which is about 2.2 degrees. Then 1.6 sin2.2 will give the elevation.

To find the angle of banking of a curved railway track, we can start by using the formula:

tan(θ) = (v^2) / (g * r)

Where:
θ is the angle of banking
v is the velocity of the train
g is the acceleration due to gravity
r is the radius of the curved track

First, let's convert the maximum safety speed limit from km/h to m/s:
54 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 15 m/s

Now, we know that the radius of the curved railway track is 600m.

Plugging these values into the formula, we have:

tan(θ) = (15^2) / (9.8 * 600)

Next, we solve for θ by taking the inverse tangent of both sides:

θ = tan^(-1) [(15^2) / (9.8 * 600)]

Using a calculator, we find that θ ≈ 18.9 degrees.

To find the elevation of the outer track above the inner track, we can use the formula:

elevation = (r * tan(θ)) - d

Where:
elevation is the height difference between the outer and inner tracks
r is the radius of the curved track
θ is the angle of banking
d is the distance between the rails

Given the radius of the track is 600m, the angle of banking (θ) is 18.9 degrees, and the distance between the rails (d) is 1.6m, we can substitute these values into the formula:

elevation = (600 * tan(18.9)) - 1.6

Using a calculator, we find that the elevation is approximately 190.2 meters. Therefore, the outer track is elevated about 190.2 meters above the inner track.