Solution.
L0=15cm=0.15m;L_0=15cm=0.15m;L0=15cm=0.15m;
T0=0oC;T_0=0^oC;T0=0oC;
T1=400oC;T_1=400^oC;T1=400oC;
αL=1.1⋅10−51oC\alpha_L=1.1\sdot10^{-5 }\dfrac{1}{oC}αL=1.1⋅10−5oC1;
αL=1LΔLΔT; ⟹ \alpha_L=\dfrac{1}{L}\dfrac{\Delta L}{\Delta T};\impliesαL=L1ΔTΔL;⟹ ΔL=αLΔTL;\Delta L=\alpha_L\Delta T L;ΔL=αLΔTL;
ΔL=1.1⋅10−51oC⋅400oC⋅0.15m=66⋅10−5m;\Delta L=1.1\sdot10^{-5}\dfrac{1}{oC}\sdot400^oC\sdot0.15m=66\sdot10^{-5}m;ΔL=1.1⋅10−5oC1⋅400oC⋅0.15m=66⋅10−5m;
L=L0+ΔL;L=L_0+\Delta L;L=L0+ΔL;
L=0.15m+0.00066m=0.15066m;L=0.15m+0.00066m=0.15066m;L=0.15m+0.00066m=0.15066m;
S=L2;S=L^2;S=L2;
S=0.15066m⋅0.15066m=0.0227m2;S=0.15066m\sdot0.15066m=0.0227m^2;S=0.15066m⋅0.15066m=0.0227m2;
Answer:S=0.0227m2.S=0.0227m^2.S=0.0227m2.
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