p140w07_ct_22

# p140w07_ct_22 - Physics 140 Winter 2007 Lecture#21 Dave...

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Unformatted text preview: Physics 140: Winter 2007 Lecture #21 March 29, 2007 Dave Winn Racquetball Striking a Wall Copyright: Loren M. Winters Mt. Etna Andrew Davidhazy Force balance and buoyancy • Submerged block BF = ρ water V block g W = ρ block V block g Σ F = BF – W Σ F=( ρ water- ρ block )V block g W BF If ρ water > ρ block : rises to surface until partly sticking out If ρ block > ρ water : sinks to bottom until partly supported by bottom If ρ water = ρ block : remains in place, neutrally buoyant Sinks: ρ water < ρ object Σ F = BF – W + F N = 0 F N = W - BF Sinks to the bottom until partly supported by the bottom Floats: ρ water > ρ object Σ F = BF – W = 0 ρ water V in g - ρ log V total g = 0 V in / V total = ρ log / ρ water Rises to the surface until part of it sticks out of the water W displaced = W floating object Tip of the iceberg…. ρ ice = 914 kg/m 3 ρ water = 1000 kg/m 3 91% of iceberg is below the surface! Neutrally buoyant: ρ fluid = ρ object Σ F = BF – W = 0 ρ fluid = ρ object A boat is floating on a lake. A large anchor is placed in the boat. The water level in the lake (with respect to the shore) 1. rises. 2. drops. 3. remains the same. Before: ρ water V displaced g= W boat After: ρ water V displaced g= W boat + W anchor A boat carrying a large anchor is floating on a lake. The anchor is thrown overboard and sinks. The water level in the lake (with respect to the shore) 1. rises....
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p140w07_ct_22 - Physics 140 Winter 2007 Lecture#21 Dave...

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