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.
Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds = 10^15 seconds
To express Picoseconds into any of other two, you have to divide 10^-12 by the power index of the one in question
1Picosecond : 10^-12 / 10^-18 = 10^ (-12- 18) = 10^ (-12+18)= 10^6 zeptoseconds
1Picosecond : 10^-12 / 10^15 = 10^ (-12-15) = 10^-27 Petaseconds.
1Picosecond = 10^6 zeptoseconds
1Picosecond = 10^-27 Petaseconds
Explanation:
Given parameters:
Mass of Neil Armstrong = 160kg
Gravitational pull of earth = 10N/kg
Moon's pull = 17% of the earth's pull
Unknown:
Difference between Armstrong's weight on moon and on earth.
Solution:
To find the weight,
Weight = mass x acceleration due to gravity = mg
Moon's gravitational pull = 17% of the earth's pull = 17% x 10 = 1.7N/kg
Weight on moon = 160 x 1.7 = 272N
Weight on earth = 160 x 10 = 1600N
The difference in weight = 1600 - 272 = 1328N
The weight of Armstrong on earth is 1328N more than on the moon.
Learn more:
Weight and mass brainly.com/question/5956881
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