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The active element of a certain laser is made of a glass rod 31.0 cm long and 1.30 cm in diameter. Assume the average coefficient of linear expansion of the glass is equal to 9.00 10-6 (°C)-1. The temperature of the rod increases by 62.0°C.
What is the increase in its length?
(a) What is the increase in its length?
(b) What is the increase in its diameter?
(c) What is the increase in its volume?

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Khoso123

Answer:

(a) ΔL=0.172 mm

(b) Δd=7.254×10⁻³mm

(c) ΔV=21.92 mm

Explanation:

Given data

L(length)=31.0cm→310mm

d (diameter)=1.30cm→13 mm

α(Coefficient)=9.00×10⁻⁶ °C⁻¹

T (Temperature)=62 °C

(a) ΔL=?

(b) Δd=?

(c) ΔV=?

Solution

For (a) ΔL part

ΔL=α×LΔT

ΔL=(9.00×10⁻⁶)×(310)×(62)

ΔL=0.172 mm

For  (b) Δd part

Area is π×r² but diameter is just 2π×r

Δd=α×dΔT

Δd=(9.00×10⁻⁶)×(13)×(62)

Δd=7.254×10⁻³mm

For (c) ΔV part

Volume is Lπ×r² and α is 3α

ΔV=3α (Lπ×r²)×62

ΔV=3×(9.00×10⁻⁶)×(310×(13/2)²) ×62

ΔV=21.92 mm

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