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Aprilv 13

# Aprilv 13 - Cory Poynter Due 22-2 11 12 21 48,55,57,59 2...

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Cory Poynter Due 04/27/09 General physics III 22-2, 11, 12, 21, 48,55,57,59 2) Mass of the gun M = 1.80kg Mass of the bullet m =2.40g Speed of the bullet u = 320m/s Energy of bullet =1.10 Ciron = 448 J/kgC = K 12mu2 = K 12 (. ) / 0012 320m s2 = . K 122 88J = . K 1 10100T T= . . 1 10100122 88J = 11170.91J Q = . 98 9100T = . . 98 910011170 91J = ( ) Q MCiron ∆T = . 1 80kg * / ( ) 448j kg℃ ∆T = . T 13 7℃ 11) Tc =373K Th =453K Qc =20kJ Weng =1.5kJ Q h = Q c + W eng = 20 kJ + 1.5 kJ = 21.5 kJ e = W eng /Q h = 1.5 kJ/21.5 kJ = 0.0698

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Cory Poynter Due 04/27/09 General physics III e C = 1 − Tc/Th = 1 – 373 K/453 K = 0.177 The actual efficiency of 0.0698 is less than four-tenths of the Carnot efficiency of 0.177. 12) Tc =500K Th =350K a) e C = 1 − Tc/Th = 1 – 350 K/500 K = 0.3 b) Th = 501K e C = 1 − Tc/Th = 1 – 350 K/501 K = 0.301 Th = 502K e C = 1 − Tc/Th = 1 – 350 K/502 K = 0.302 Th = 503K e C = 1 − Tc/Th = 1 – 350 K/503 K = 0.304 c) Tc = 349K e C = 1 − Tc/Th = 1 – 349 K/500 K = 0.302
Cory Poynter Due 04/27/09 General physics III Tc = 348K e C = 1 − Tc/Th = 1 – 348 K/500 K = 0.304 Tc = 347K e C = 1 − Tc/Th = 1 – 347 K/500 K = 0.306 21) The energy drawn from the cold reservior Qc The energy drawn from the cold reservior Qh a) = WQc = Q h- QcQc = Q h Qc - 1 Q h = Qc T h Tc = WQc T h Tc - 1 = W Tk - TcTcQc

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