Aspherical steel ball bearing has a diameter of 2.540 cm at 20.00°c. (assume the coefficient of linear expansion for steel is 11 ✕ 10−6 (°c)−1. ) (a) what is its diameter when its temperature is raised to 87.0°c? (give your answer to at least four significant figures.) 2.542 correct: your answer is correct. cm (b) what temperature change is required to increase its volume by 0.900%

(a) 2.542 cm

(b) 272.7°C

Explanation:

diameter, d = 2.540 cm

T1 = 20°C

α = 11 x 10^-6 /°C

(a) Let d' be the diameter.

T2 = 87°C

Use he formula for the areal expansion

A' = A ( 1 + βΔT)

where, β is the coefficient of areal expansion and ΔT is teh rise in temperature, A' be the area at high temperature and A be the area at low temperature.

β = 2 α = 2 x 11 x 106-6 = 22 x 10^-6 /°C

So,

D'^2 = 2.54^2 ( 1 + 22 x 10^-6 x 67)

D' = 2.542 cm

(b) Let the change in temperature is ΔT.

Use the formula for the volumetric expansion

ΔV = V x γ x ΔT

Where, γ = 3 x α = 3 x 11 x 10^-6 = 33 x 10^-6 /°C

0.9/100 = 33 x 10^-6 x ΔT

ΔT = 272.7°C

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

the area between the graph and the t axis