<span>Kepler's second law: as a planet moves around its orbit, it sweeps out EQUAL areas in EQUAL times.
</span>
Answer:
- the coating’s index of refraction is 1.25
- the required thickness is 104.1667 nm
Explanation:
Given the data in the question;
Thickness of coating t = 100 nm
wavelength λ = 500nm
we know that refractive index is;
t = λ/4n
make n, the subject of formula
t4n = λ
n = λ / 4t
we substitute
n = 500 / ( 4 × 100 )
n = 500 / 400
n = 1.25
Therefore, the coating’s index of refraction is 1.25
2)
given that;
Index of refraction of the coating; n = 1.20
λ = 500 nm
thickness of coating t = ?
t = λ / 4n
we substitute
t = 500 / ( 4 × 1.2 )
t = 500 / 4.8
t = 104.1667 nm
Therefore, the required thickness is 104.1667 nm
Mgh= 1/2 m v^2
gh= 1/2 v^2 9.8 * 8= 1/2 v^2
solve for v
V = sqrt(2gH)where H = 8 m.
Scale: (soft 1-->6 hardest)
<span>1=Talc 2=Gypsum 3=Calcite 4=Fluorite 5=Apatite 6=Orthoclase
</span>Mineral #1 can only scratch two other minerals therefore it must have a hardness level of 2+1=3 which is Calcite. (scratches talc, gypsum)
Mineral #2 can scratch four other minerals therefore it must have a hardness level of 4+1=5 which is Apatite. (scratches all but Apatite, Orthoclase)
Looking through the possible conclusions.. It looks like answer is D.
Answer: Radiation
Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.
Examples of radiation include the light, heat, and particles emitted from the Sun.
Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.
So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.
