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anzhelika [568]
3 years ago
11

Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has

a bulk modulus of 2.3 × 109 N/m2. Bulk modulus is defined to be
B ≡ rho0 ∆P ∆rhoAnswer in units of kg/m3.
Physics
1 answer:
Pie3 years ago
4 0

Answer:

1060.41kg/m^3

Explanation:

Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:

B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}

Hence the density of the seawater at a depth of 680atm is calculated as:-

\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

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V_d = 1.75 × 10⁻⁴ m/s

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Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

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Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

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