The answer to this item is rondo, letter A. As already mentioned in the above's definition, this has a recurring or repetitive lead them two contrasting sections. The contrasting sections are more commonly referred to as "episodes" and occasionally as "digressions" or "couplets".
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.
Assuming that the angle is the same for both ropes, then D. is the answer. You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
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When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
To know more about angular magnification refer:: brainly.com/question/28325488
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