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8 days ago

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If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of th

e hole when the rivet is cooled to - 78.0 ∘C, the temperature of dry ice? Assume that the expansion coefficient of aluminum is 2.4×10−5(∘C)−1 and remains constant.

1 answer:

kicyunya [2.9K]8 days ago

8 0

Answer:

Explanation:

at 23 degrees Celsius, the diameter measures 4.511 mm

GIVEN DATA:

diameter of hole = 4.500 mm

T_1 = 23.0 degrees Celsius

T_2 = - 78.0 degrees Celsius

the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}

the diameter at 23 degrees Celsius is stated as

$d = d_o (1+ \alpha \delta T)$

$= 4.5 (1+2.4*10^{-5} *(23-(-78)))$

= 4.511 mm

the diameter of the rivet after temperature change is given as

$d= d_o + \Delta d$

$= d_o(1+ \alpha \delta T)$

$= 0.4500 *(1+2.4*10^{-5} *(23-(-78)))$

= 0.4511 cm

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