Answer:
Δ
= 84 Ω, = (40 ± 8) 10¹ Ω
Explanation:
The formula for parallel equivalent resistance is
1 / = ∑ 1 / Ri
In our case we use a resistance of each
R₁ = 500 ± 50 Ω
R₂ = 2000 ± 5%
This percentage equals
0.05 = ΔR₂ / R₂
ΔR₂ = 0.05 R₂
ΔR₂ = 0.05 2000 = 100 Ω
We write the resistance
R₂ = 2000 ± 100 Ω
We apply the initial formula
1 / = 1 / R₁ + 1 / R₂
1 / = 1/500 + 1/2000 = 0.0025
= 400 Ω
Let's look for the error (uncertainly) of Re
= R₁R₂ / (R₁ + R₂)
R’= R₁ + R₂
= R₁R₂ / R’
Let's look for the uncertainty of this equation
Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’
The uncertainty of a sum is
ΔR’= ΔR₁ + ΔR₂
We substitute the values
Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)
Δ / 400 = 0.1 + 0.05 + 0.06
Δ = 0.21 400
Δ = 84 Ω
Let's write the resistance value with the correct significant figures
= (40 ± 8) 10¹ Ω
Answer: F = 20 N
Explanation:
I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.
To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)
The easiest convenient point is the fulcrum
30((90/2) - 15) - F(90/2) = 0
30(30) = F(45)
F = 900/45 = 20 N
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I dont know the answer sorry but id did read it wrong at first and i thought it said a sick person was thrown off a cliff LOL
