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Given the data, the correct statement is
Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s
<h3>What is acceleration? </h3>This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 15 m/s
- Time (t) = 10 s
- Initial acceleration (a₁) =?
a₁ = (v – u) / t
a₁ = (15 – 0) / 10
a₁ = 1.5 m/s²
<h3>How to determine the final acceleration of Beverly</h3>- Initial velocity (u) = 15 m/s
- Final velocity (v) = 35 m/s
- Time (t) = 50 - 10 = 40 s
- Final acceleration (a₂) =?
a₂ = (v – u) / t
a₂ = (35 – 15) / 40
a₂ = 0.5 m/s²
<h3>How to determine the initial acceleration of Carl</h3>- Initial velocity (u) = 0 m/s
- Final velocity (v) = 10 m/s
- Time (t) = 10 s
- Initial acceleration (a₁) =?
a₁ = (v – u) / t
a₁ = (10 – 0) / 10
a₁ = 1 m/s²
<h3>How to determine the final acceleration of Carl</h3>- Initial velocity (u) = 10 m/s
- Final velocity (v) = 30 m/s
- Time (t) = 50 - 10 = 40 s
- Final acceleration (a₂) =?
a₂ = (v – u) / t
a₂ = (30 – 10) / 40
a₂ = 0.5 m/s²
SUMMARY
- Initial acceleration of Beverly = 1.5 m/s²
- Final acceleration of Beverly = 0.5 m/s²
- Initial acceleration of Carl = 1 m/s²
- Final acceleration of Carl = 0.5 m/s²
From the above calculations, we can see that Beverly's initial acceleration is higher than that of Carl's and their final acceleration is the same.
Therefore, the correct answer to the question is:
Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s
Complete question
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First, we write out a balanced equation.
HA <--> H(+) + A(-)
Next, we create an ICE table
HA <--> H+ + A-
[]i 0.40M 0M 0M
Δ[] -x +x +x
[]f 0.40-x x x
Next, we write out the Ka expression.
Ka = [H+][A-]/[HA]
Ka = x*x/(0.40-x)
However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,
Assume 0.40-x ≈ 0.40
Therefore, 1.2x10^-6 = x^2/0.40
Then we solve for the [H+] concentration, or x
x=6.93x10^-4
Next, to find pH we do
pH = -log[H+]
pH = -log[6.93x10^-4]
pH = 3.2
HA <--> H(+) + A(-)
Next, we create an ICE table
HA <--> H+ + A-
[]i 0.40M 0M 0M
Δ[] -x +x +x
[]f 0.40-x x x
Next, we write out the Ka expression.
Ka = [H+][A-]/[HA]
Ka = x*x/(0.40-x)
However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,
Assume 0.40-x ≈ 0.40
Therefore, 1.2x10^-6 = x^2/0.40
Then we solve for the [H+] concentration, or x
x=6.93x10^-4
Next, to find pH we do
pH = -log[H+]
pH = -log[6.93x10^-4]
pH = 3.2