Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.
hope it helps
Core
Home of atoms of hydrogen also the lightest element in the universe.
Radiative Zone
Outside the inner Core it radiates energy through the process of photon emission.
Convection Layer
Outer most Layer of the Core, it extends form a depth of 200,000 kilometres to the visible surface. Energy is created by Convection. This is where light is produced.
Photosphere
Surrounds the stars and is where light and heat radiate.
Chromosphere
Reddish gas layer outside of the photosphere I think it also works with the Corona.
Corona
Aura of Plasma that surrounds the Sun and other stars, it extends millions of kilometres and easily seen during a total eclipse.
1.) The object's Velocity
Faster it goes, more kinetic energy it has
From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one.