Physics Question on Bouyancy *EASY 10 POINTS*
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Physics Question on Bouyancy *EASY 10 POINTS*

[From: ] [author: ] [Date: 13-08-11] [Hit: ]
so we need to round our result to 2 sig digs, which gives: (8) V2 / V1 = 1.0So the gold bars contraction in volume due to undersea pressure is negligible (nugatory)........

β = 1 / 180GPa.

Now we need to find the pressure at 3.9km under the ocean surface, which will be our Δp: Hydrostatic pressure p at depth h below the surface is given by the formula (see Source 3):

(6) p = p₀ + ρ∙g∙h,

where:
p₀ is the atmospheric pressure at water surface, usually p₀ = 1atm = 101325Pa
ρ is the density of seawater, given as 1050 kg/m^3
g is gravitational acceleration = 9.81m/s^2
h is the depth = 3.9e3 m

(Since p₀ << ρ∙g∙h, I'm going to ignore it)

Substituting these into (5) gives (7):

(7) V2 / V1 = (1 - β * Δp)

= (1 - ( 1 / 180GPa) * (ρ * g * h) )

= (1 - ρ * g * h / 180GPa )

= 1 - 1050 * 9.81 * 3.9e3 / 180e9

= 0.9997

Our least precise data value has only 2 significant digits (3.9km), so we need to round our result to 2 sig digs, which gives:

(8) V2 / V1 = 1.0

So the gold bar's contraction in volume due to undersea pressure is negligible (nugatory).

.
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