A train travels 650 meters in 25 seconds. What is the train's velocity?
1 answer:
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Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Answer:
decreases by 1/9
Explanation:
the equation of kinetic energy is
here, we see that Kinetic Energy and v (velocity) are on opposide sides of the equation, this means that they are directly proportional (when one increases, the other one must increase (assuming mass (m) is constant)): KE∝
since V is decreasing by factor of 1/3, simply plug it in for v^2 to get (1/3)^2,
this equals (1/9). So KE will become 1/9 of original value