Sebuah pelat alumunium dngn luas 0,5 m² dipanaskan hingga suhunya naik sebesar 90 C. Luas pelat alumunium setelah dipnskan adalah... m². (α=0,0000255/C)
tlg dijawab.. pakai diketahui dan rumusya yg jelas...
udinhoirDik : A(mula2) = 0,5 m² α = 225 x 10 ⁻⁷ / C diproleh β = 450 x 10 ⁻⁷/ C Δt = 90 °C Dit: A(akhir)? A (akhir) = A (mula2) . (1 + β . Δt) = ( 0,5) . (1 + 450 x 10 ⁻⁷ . 90) = 0,5 + 20250 x 10 ⁻⁷ = 0,502025 m²
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lilisdwiastri1
maaf.. yg koefisienny 0,0000255/C bukan 225
Aderachma0313
Dik = A₀ = 0,5 m² Δt = 90°C α = 2,55×10⁻⁵/°C β = 2.α β = 5,1×10⁻⁵/°C Dit = At Jwb = At = A₀.(1+β.Δt) At = 0,5.(1 + 5,1×10⁻⁵.90) At = 0,5.(1+4,59×10⁻³) At = 0,5+2,295×10⁻³ At = 0,502295 m² At = 0,502 m²
α = 225 x 10 ⁻⁷ / C diproleh β = 450 x 10 ⁻⁷/ C
Δt = 90 °C
Dit: A(akhir)?
A (akhir) = A (mula2) . (1 + β . Δt)
= ( 0,5) . (1 + 450 x 10 ⁻⁷ . 90)
= 0,5 + 20250 x 10 ⁻⁷
= 0,502025 m²
A₀ = 0,5 m²
Δt = 90°C
α = 2,55×10⁻⁵/°C
β = 2.α
β = 5,1×10⁻⁵/°C
Dit = At
Jwb =
At = A₀.(1+β.Δt)
At = 0,5.(1 + 5,1×10⁻⁵.90)
At = 0,5.(1+4,59×10⁻³)
At = 0,5+2,295×10⁻³
At = 0,502295 m²
At = 0,502 m²