What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el
1 answer:
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Answer:
<em>0.85c </em>
Explanation:
Rest mass of Kaon = 494 MeV/c²
Rest mass of proton = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy = 494c² MeV
for the proton, rest energy = 938c² MeV
Recall that the rest energy, and the total energy are related by..
= γ
which can be written in this case as
= γ
...... equ 1
where = total energy of the kaon, and
= rest energy of the kaon
γ = relativistic factor =
where
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
=
......equ 2
where is the total energy of the kaon, and
is the rest energy of the proton.
From =
= 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = = 1.89
1.89 = 1
squaring both sides, we get
3.57( 1 - ) = 1
3.57 - 3.57 = 1
2.57 = 3.57
= 2.57/3.57 = 0.72
= 0.85
but,
v/c = 0.85
v = <em>0.85c </em>
Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
(a) the electrical power generated for still summer day
(b)the electrical power generated for a breezy winter day