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.
Answer:
0.0319 m³
Explanation:
Use ideal gas law:
PV = nRT
where P is pressure, V is volume, n is amount of gas, R is the gas constant, and T is temperature.
Since P, n, and R are held constant:
n₁ R / P₁ = n₂ R₂ / P₂
Which means:
V₁ / T₁ = V₂ / T₂
Plugging in:
0.0279 m³ / 280 K = V / 320 K
V = 0.0319 m³
Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the change the object's motion.
Answer:
She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.
Explanation:
mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.
Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.
