PHYS211 LAB 13. THERMAL EXPANSION 8. Suppose on a typical spring/fall day (15.0°

Question

PHYS211 LAB 13. THERMAL EXPANSION 8. Suppose on a typical spring/fall day (15.0°C) you measure a e on a typical spring/fall day (15.0°C) you measure a PVC pipe to be 3.0480 m a steel measuring tape. What will be the length of the pipe on a cold winter ) or very hot summer day (40.0°C)? Remember to take into account the long with day (-10.0°C) or very hot summer day (40.0 fact that the steel tape will also undergo thermal expansion/contraction Suppose a railroad company decides to go with continuous welded rail to eliminate the typical clackity clack from 12m long jointed tracks. Suppose the rail is not wel aftixed to the rail bed, except at the ends, and thus is free to expand. Suppose further it is flat on the ground at 0.0°C. Find first the expanded length of L = 1500, m of steel rail at 40.0°C. Then, use this length (L+ AL) as the hypothenuses of two right triangles with common legs with the other legs being the unexpanded length. Find the length of the common legs (h = V(750+AL72)2.7502). If the two rails went in opposite directions, is this enough to derail a train? L = )m

Explanation / Answer

Coefficient for linear expansion of steel a = 12*10-6 deg C-1

Final Length L = L0 ( 1 + aT) where L0 is the initial length and T is the temperature change

L-10 = 3.0480 (1 + 12 * 10-6 * -25) = 3.0470m

1m on tape will now become L = 1 ( 1 + 12 * 10-6 * -25) = 0.9997m

So L-10 = 3.0470/0.9997 = 3.0479m

L40 = 3.0480 (1 + 12 * 10-6 * 25) = 3.0489m

1m on tape will now become L = 1 ( 1 + 12 * 10-6 * 25) = 1.0003m

So L40 = 3.0489/1.0003 = 3.0480m

2) del L = a L T =   12 * 10-6 * 1500 * 40 = 0.72 m

h = ((750 + del L /2)2 - 7502)1/2 = 32.87m

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