Answer:
Yes
Explanation:
Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.
<u>That is given by as relation:</u>
where:
absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation
reflectivity is defined as the ratio of reflected radiation to the total irradiation
transmittivity is defined as the ratio of total transmitted radiation to the total irradiation
Answer:
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Explanation:
In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.
On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.
Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.
Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:
Metal 1
ρ = m/V
ρ = 22.1 g / 3 cm³
ρ = 7.367 g / cm³
Metal 2
ρ = m/V
ρ = 42 g / 4 cm³
ρ = 10.5 g / cm³
Metal 3
ρ = m/V
ρ = 9.32 g / 5 cm³
ρ = 1.864 g / cm³
Metal 4
ρ = m/V
ρ = 8.13 g / 3 cm³
ρ = 2.71 g / cm³
<h2>R / Metal 4 was selected.</h2>
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
Troposphere is the answer
