A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?

the final temperature is

Explanation:

The change in length of a bar can be expressed with the relation;

  ---- (1)

Also ; the relative or fractional increase in length  is proportional to the change in temperature.

Mathematically;

ΔL/L_i ∝ kΔT

where;

k is replaced with ∝ (the proportionality constant )

   ---- (2)

From (1) ;

 ---  (3)

from (2)

 ---- (4)

replacing (4) into (3);we have;

On re-arrangement; we have

from the given question; we can say that :

So;

Making the change in temperature the subject of the formula; we have:

where;

= 10.02 cm

= 10.00 cm

= 19 × 10⁻⁶ °C ⁻¹

= 24  × 10⁻⁶ °C ⁻¹

= −396.1965135 ° C

≅ −396.20  °C

Given that the initial temperature

Then ;

Thus;

Thus; the final temperature is

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y