Answer:
The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%
Explanation:
Let V be the total volume of the iceberg
Let x be the volume of iceberg submerged
According to Archimedes principle,
weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)
Weight of the iceberg = mg= ρ(iceberg) × V × g
Weight of water displaced = ρ(fluid) × x × g
We then have
ρ(iceberg) × V × g = ρ(fluid) × x × g
(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%
Hope this Helps!!!!
Q. The energy emitted from the sun is a product of ________.
A. Fusion
Magnetic fields
Explanation:
The presence of magnetic fields best explains why a magnet can act a distance on other magnets or on objects containing certain metals.
- Magnetic fields are lines of forces around a bar magnet.
- These lines of forces attracts and repels other magnetic bodies and metallic bodies round it.
- Magnetic lines of forces originates at the north pole and enters in the south pole.
- Areas around a magnetic body are bounded by force fields.
- A magnet has permanent magnetic fields round it.
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Answer:
The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.
Explanation:
Equipotential surface is a three dimensional part of equipotential lines.
Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.
Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.
Answer:
θ = 1.591 10⁻² rad
Explanation:
For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source
Let's write the diffraction equation for a slit
a sin θ = m λ
The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ
θ = λ / a
In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.
θ = 1.22 λ / D
Where D is the diameter of the pupil
Let's apply this equation to our case
θ = 1.22 600 10⁻⁹ / 0.460 10⁻²
θ = 1.591 10⁻² rad
This is the angle separation to solve the two light sources
