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If the observer is moving away from the source ((Figure)), the observed frequency can be found: λs=vTo−voTovTs=(v−vo)Tov(1fs)=(v−vo)(1fo)fo=fs(v−vov).
A. reactants
B. subscript
C. coefficient
D. products
Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.
Size and mass have no effect.</span>
(186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)
= (186,000 x 3,600 x 24 x 365) mi/yr
= 5,865,696,000,000 miles per year (rounded to the nearest million miles)
