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valentina_108 [34]
3 years ago
7

Older railroad tracks in the U.S. are made of 12-m-long pieces of steel. When the tracks are laid, gaps are left between section

s to prevent buckling when the steel termally expands. If a track is laid at 16*C, how large should the gaps be if the track is not to buckle when the temperature is as high as 50*C?
Physics
2 answers:
never [62]3 years ago
8 0

Answer: gap required in track =

5.39millimeter{5.39mm}

Explanation: This is a linear expansivity problem.

Coefficient of Linear expansivity = {gap in lenght}/{{original lenght*.

{change in temperature}}

But coefficient of linear expansivity of steel = 13.2*EXP{-6}/degree Celsius.

Making change in lenght or gap subject of formula we have,

Gap = coefficient of linear expansivity*{{original lenght*{change in temperature}}

But,

Original lenght = 12m

Change in temperature = 50 - 16 = 34 degree Celsius.

Therefore,

Gap in lenght = 13.2*EXP{-6}*12*34=

=5.39*EXP{-3}meter

= 5.39mm

____ [38]3 years ago
6 0

Answer: ∆L = 0.49cm ≈ 0.50cm

Therefore there should be 0.5 cm gap between each piece of steel.

Explanation:

Thermal expansion of steel is the increase in size of steel as a result of increased temperature. It can be represented by the mathematical expression:

∆L = L(k)∆T .....1

Where;

∆L is the change in length

L is the initial length

∆T is the change in temperature

k is the specific Linear expansion coefficient.

Given;

L = 12m

∆T = 50°C - 16°C = 34°C

k (for steel) = 1.2 × 10^-6 /C

Substituting the values into the equation 1

∆L = 12 × 34 × 12×10^-6

∆L = 4896 × 10^-6 m

∆L = 0.49cm ≈ 0.50cm

Therefore there should be 0.5 cm gap between each piece of steel.

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8 0
2 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

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r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

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B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

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3 years ago
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| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

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                                      11.1 MHz           <== online impedance calculator

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