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Vera_Pavlovna [14]
3 years ago
6

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

ansion joints will be provided to compensate for the change in length in the girders as the temperature varies. Each expansion joint can compensate for a maximum of 20 mm of change in length. From historical records it is estimated that the minimum and maximum temperatures in the region will be -35°C and 40°C, respectively. Determine (a) the minimum number of expansion joints required and (b) the length that each bridge section should be fabricated. The coefficient of thermal expansion α=12(10-6)/C for steel.
Physics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

23

21.7391304348 m

Explanation:

L = Initial length = 500 m

\Delta T = Change in temperature = 40-(-35)

\alpha = Coefficient of thermal expansion = 12\times 10^{-6}\ /^{\circ}C

Change in length is given by

\Delta L=L\alpha \Delta T\\\Rightarrow \Delta L=500\times 12\times 10^{-6}\times (40-(-35))\\\Rightarrow \Delta L=0.45\ m

The change in length is 0.45 m

The number of joints would be

n=\dfrac{0.45}{0.02}\\\Rightarrow n=22.5\\\Rightarrow n\approx 23

The number of joints is 23

Each bridge section length would be

\dfrac{L}{n}=\dfrac{500}{23}=21.7391304348\ m

The length of each bridge section would be 21.7391304348 m

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A 295-kg object and a 595-kg object are separated by 4.10 m.
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a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

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The force between the mass m₁ and m₃

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by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

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F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
3 years ago
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