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For this problem, first, lets recabe information:
v = 0 m/s
t = 3 s
g = 9,82 m/s²
v' = ?
d = ?
First, for calculate the final velocity:
v' = v + g * t
v' = 0 m/s + 9,82 m/s² * 3 s
v' = 29,46 m/s
Now, for calculate how far did it fall:
d = v * t + g * t^2 / 2
Like v = 0 m/s, we can simplificate equation:
d = g * t^2 / 2
d = 9,82 m/s² * (3 s)^2 / 2
d = 9,82 m/s² * 9 s² / 2
d = 88,38 m / 2
d = 44.19 m
¡Buena suerte!
Let R₁ and R₂ be the two Resistance
R₁ - Resistance of 1st resistor
R₂ - Resistance of 2nd resistor
V = Voltage = 12V
I = O.31A ( in series)
I = 1.6 A ( in parallel)
when two resistance connected in series
Rs = R₁ + R₂
V = IRs = 12 /0.31 V
R₁+R₂ = 38.700Ω (equation1.)
When two resistance connected in parallel
V = I Rp
V= I (R₂ R₁ /R₁ +R₂)
R₁ R₂ = 290.25 Ω
As we know
(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2
Using above values we get,
(R₁ - R₂ ) ² = (38.70) ² - 4x 290.25
(R₁ - R₂ ) ² = 336.69 Ω
R₁ - R₂ = 18.34 (equation 2 )
Using equation 1 and 2 we get
R₁+ R₂ = 38.70 52
R₁ - R₂ = 18.34
R₁ = 28.535 Ω
Using the value of R₁ in equation 1 we get
R₂ = 38·700 - 28.535
RR₂ = 10.125 Ω
R₁ = 28. 535 v and R₂ 10.125 Ω
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