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:
3.066×10^21 photons/(s.m^2)
Explanation:
The power per area is:
Power/A = (# of photons /t /A)×(energy / photon)
E/photons = h×c/(λ)
photons /t /A = (Power/A)×λ /(h×c)
photons /t /A = (P/A)×λ/(hc)
photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)
= 3.066×10^21
Therefore, the number of photons per second per square meter 3.066×10^21 photons/(s.m^2).
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Answer:
F = f from Newton’s first law.
Explanation:
since the desk is moved in a straight line at a constant speed, newton first law tell us that the two forces must be equal.
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. since the table has been set in motion by the 400 N force, it will remain in motion unless it is been acted upon by an external force, and this means that the 400 N must be equal to the frictional force for it to have been in motion in the first instance.
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
- W = F║d
- where the force is parallel to the displacement
Looking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
- W = (87 N)(4.5 m)
- W = 391.5 J
The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
