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Unformatted text preview: is the sum of the pressure due to the weight of the water plus the atmospheric pressure pushing on the surface of the water. F from_outside = (h·g· ρ water + p atmosphere )·A Door 8) F from_inside = p atmosphere ·A Door 9) Conservation of energy: ½ m v y 2 = mgh b v y 2 = 2gh v = sqrt(v x 2 +v y 2 ) = sqrt(v x 2 +2gh) ; sqrt: ‘square root’ h y x v x v y v h v x 10) At maximum load: F net = 0 = F Buoancy – g( m Balloon + m Helium + m cargo ) b m cargo = F Buoancy /g  m Balloon – m Helium The bouyant force is equal to the weight of the displaced air: F Buoancy = V Baloon · ρ Air = 4/3 π r 3 m Helium = V Baloon · ρ Helium 1113) y = A · sin(kx – ω t) k = 2 π / λ , λ : ‘wavelength’ ω = 2 π /T , T 0 : ‘Period’ 14) v y = dy/dt = A· ω ·cos(kx – ω t) b max vertical speed: v y,max = A· ω...
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This note was uploaded on 11/07/2010 for the course PHYS 1120 taught by Professor Rogers during the Spring '08 term at Colorado.
 Spring '08
 ROGERS
 Physics, Mass

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