A closed right cylindrical tank with a height of 8 m and 3 m in diameter contains water to a depth of 5m. If the tank is laid in a horizontal position, find the depth of water.
cos12θ=0.5/2.5cos\frac{1}{2}θ=0.5/2.5cos21θ=0.5/2.5
θ=156.926∘θ=156.926^∘θ=156.926∘
α=360∘−θ=360∘−156.926∘α=360^∘−θ=360^∘−156.926^∘α=360∘−θ=360∘−156.926∘
α=203.074∘α=203.074^∘α=203.074∘
Area of cross section:
Ab=1/2r2(αrad+sinθdeg)A_b=1/2r^2(α_{rad}+sinθ_{deg})Ab=1/2r2(αrad+sinθdeg)
Ab=12.3ft2A_b=12.3 ft^2Ab=12.3ft2
Volume of water:
Vw=AbL=12.3(8)V_w=A_bL=12.3(8)Vw=AbL=12.3(8)
Vw=98.4V_w=98.4Vw=98.4
Depth of water in vertical position:
Vw=πr2hVw=πr^2hVw=πr2h
98.4=π(2.52)h98.4=π(2.5^2)h98.4=π(2.52)h
h=5.01 ft answerh=5.01\space ft \space\space answerh=5.01 ft answer
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