Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
1 answer:
Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt =
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
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Explanation:
1) N₂ + O₂ → 2 NO
Kc = [NO]² / ([N₂] [O₂])
Set up an ICE table:
Plug into the equilibrium equation and solve for x.
1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))
1.00×10⁻⁵ = (2x)² / (0.114 − x)²
√(1.00×10⁻⁵) = 2x / (0.114 − x)
0.00316 = 2x / (0.114 − x)
0.00361 − 0.00316x = 2x
0.00361 = 2.00316x
x = 0.00018
The volume is 1.00 L, so the concentrations at equilibrium are:
[N₂] = 0.114 − x = 0.11382
[O₂] = 0.114 − x = 0.11382
[NO] = 2x = 0.00036
2(a) Cl₂ → 2 Cl
Kc = [Cl]² / [Cl₂]
1.2×10⁻⁷ = (2x)² / (2 − x)
1.2×10⁻⁷ (2 − x) = 4x²
2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²
2.4×10⁻⁷ ≈ 4x²
x² ≈ 6×10⁻⁸
x ≈ 0.000245
2x ≈ 0.00049
2(b) F₂ → 2 F
Kc = [F]² / [F₂]
1.2×10⁻⁴ = (2x)² / (2 − x)
1.2×10⁻⁴ (2 − x) = 4x²
2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²
2.4×10⁻⁴ ≈ 4x²
x² ≈ 6×10⁻⁵
x ≈ 0.00775
2x ≈ 0.0155
F₂ dissociates more, so Cl₂ is more stable at 1000 K.