Pap Hst

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What mass of air is inside the cylinder? Solution: π Fs = ks∆x = ks ∆V/Ap ; V1 = 1 L = 0.001 m3, Ap = 4 0.12 = 0.007854 m2 State 2: V3 = 1.5 L = 0.0015 m3; T3 = 80°C = 353.15 K The pressure varies linearly with volume seen from a force balance as: PAp = P0 Ap + mp g + ks(V - V0)/Ap Between the states 1 and 2 only volume varies so: ks(V3-V2) 80×103(0.0015 - 0.001) P3 = P2 + = 150 + Ap2 0.0078542 × 1000 = 798.5 kPa P3V3 798.5 × 0.0015 m = RT = = 0.012 kg 0.287 × 353.15 3 P 3 2 1 v Sonntag,

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Mata Atlântica Echinodorus gabrielii Rataj Referência: Rataj, K. 1990. Akvár. Terár., 33: 15. Distribuição: nativa; não endêmica; Norte (PA); Amazônia Catálogo de plantas e fungos do Brasil Echinodorus glandulosus Rataj Voucher: Félix, L.P. 8996 (HST) Distribuição: nativa; endêmica; Nordeste (CE, PB, PE, BA); Caatinga Echinodorus glaucus Rataj Voucher: Silva, J.B.F. 411 (SP) Distribuição: nativa; não endêmica; Norte, Centro-Oeste (MT, MS); Amazônia, Pantanal Echinodorus grandiflorus…

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