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:
b)
Explanation:
By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.
This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.
(As the positive charge would move away from positive charges and would be attracted by negative ones).
So, the combination of answers that is true is b) (positive, negative, positive).
Answer:
option a is correct
Explanation:
<h2>I hope it's help you ❣️❣️</h2>
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ = 
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
200 Hz = 200 cycles per sec
<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>
