Suppose that at room temperature, a certain aluminum bar is 1. m long. the bar gets longer when its temperature is raised. the length l of the bar obeys the following relation: l=1.+2.4×10−5t, where t is the number of degrees celsius above room temperature. what is the change of the bar's length if the temperature is raised to 11.3 ∘c above room temperature?
express your answer in meters to two significant figures.

0.00027m

Explanation:

Given the initial length of aluminum bar Lo = 1.0000m

After heating the aluminum bar under a temperate above room temperature which is 11.3°C, its new length is modelled by the equation:

Lf =1.0000+2.4×10^−5T, where T is the number of degrees Celsius above room temperature.

If T = 11.3°C

Lf = 1.0000+(2.4×10^-5)(11.3)

Lf = 1.0000+27.12×10^-5

Lf = 1.0000+0.0002712

Lf = 1.0002712

The change in bar length = Final length Lf - initial length Lo

Change in bar length = 1.0002712-1.0000

Change in bar length = 0.0002712m

= 0.00027m (to 2s.f)

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6