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Question Number 151823 by DELETED last updated on 23/Aug/21

Answered by DELETED last updated on 23/Aug/21

54. R_1 :R_2 =ρ(l_1 /(πR_1 ^2 )):ρ(l_2 /(πR_2 ^2 )) R_1 :R_2 =(1/D_1 ^2 ) : (1/D_2 ^2 ) dan D_2 =(D_1 /2) 2 : R_2 = (1/D_1 ^2 ) : (4/D_1 ^2 ) 2:R_2 =1:4 →R_2 =8 Ω

$$\mathrm{54}.\:\:\mathrm{R}_{\mathrm{1}} :\mathrm{R}_{\mathrm{2}} =\rho\frac{\mathrm{l}_{\mathrm{1}} }{\pi\mathrm{R}_{\mathrm{1}} ^{\mathrm{2}} }:\rho\frac{\mathrm{l}_{\mathrm{2}} }{\pi\mathrm{R}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\mathrm{R}_{\mathrm{1}} :\mathrm{R}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{D}_{\mathrm{1}} ^{\mathrm{2}} }\::\:\frac{\mathrm{1}}{\mathrm{D}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:\mathrm{dan}\:\mathrm{D}_{\mathrm{2}} =\frac{\mathrm{D}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\:\:\mathrm{2}\::\:\mathrm{R}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{D}_{\mathrm{1}} ^{\mathrm{2}} }\::\:\frac{\mathrm{4}}{\mathrm{D}_{\mathrm{1}} ^{\mathrm{2}} }\: \\ $$$$\:\:\mathrm{2}:\mathrm{R}_{\mathrm{2}} =\mathrm{1}:\mathrm{4}\:\rightarrow\mathrm{R}_{\mathrm{2}} =\mathrm{8}\:\Omega \\ $$

Answered by DELETED last updated on 23/Aug/21

55). R_t =R(1+α.ΔT) 75=50(1+3,9×10^(−3) .ΔT) ((75)/(50))=1+3,9×10^(−3) .ΔT 1,5−1=3,9×10^(−3) .ΔT ΔT=((0,5)/(3,9×10^(−3) )) =((5000)/(39))=128,2 °C →Titik lebur=128,2+20=148,2 °C

$$\left.\mathrm{55}\right).\:\mathrm{R}_{\mathrm{t}} =\mathrm{R}\left(\mathrm{1}+\alpha.\Delta\mathrm{T}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{75}=\mathrm{50}\left(\mathrm{1}+\mathrm{3},\mathrm{9}×\mathrm{10}^{−\mathrm{3}} .\Delta\mathrm{T}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{75}}{\mathrm{50}}=\mathrm{1}+\mathrm{3},\mathrm{9}×\mathrm{10}^{−\mathrm{3}} .\Delta\mathrm{T} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1},\mathrm{5}−\mathrm{1}=\mathrm{3},\mathrm{9}×\mathrm{10}^{−\mathrm{3}} .\Delta\mathrm{T} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Delta\mathrm{T}=\frac{\mathrm{0},\mathrm{5}}{\mathrm{3},\mathrm{9}×\mathrm{10}^{−\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5000}}{\mathrm{39}}=\mathrm{128},\mathrm{2}\:°\mathrm{C} \\ $$$$\rightarrow\mathrm{Titik}\:\mathrm{lebur}=\mathrm{128},\mathrm{2}+\mathrm{20}=\mathrm{148},\mathrm{2}\:°\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

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