Help ASAP!!

Beverly and Carl are in a race. Their graphs show the data.

Nikitich [7]

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

<h3>How to determine the initial acceleration of Beverly</h3>

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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Learn more about acceleration:

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8 0

2 years ago

A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac

UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

$pH = pKa + log(\frac{[KF]}{[HF]})$

$pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04$

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

$\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol$

$\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles$

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

$\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles$

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)

The number of moles of HF after the reaction of KF with HCl is:

$\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles$

And the concentration of HF after the reaction of KF with HCl is is:

$C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L$

Now, from the equilibrium of equation (1) we have:

$Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}$

$Ka = \frac{x^{2}}{0.531 - x}$ (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

$pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73$

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0

4 years ago

Dr. von Waldner conducts research on depression. His hypothesis is that depression results from maladaptive thought patterns. Fr

Nimfa-mama [501]

The answer is C- cognitive

3 0

3 years ago

Na-23 has 12 neutrons. What is it’s atomic number?

erastovalidia [21]

Answer:

11

Explanation:

The atomic number = number of protons. The mass number (23) = sum of number of neutrons and protons. Since you know the mass number is 23 and it has 12 neutrons, 23-12 gives you 11 protons, so its atomic number is 11.

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6 0

3 years ago

Read 2 more answers

How do you prepare 500 ml of a 1.77M H2SO4 solution from an 18.0 M H2SO4 stock solution?

finlep [7]

Answer:

Molarity Problems Worksheet

M=nV n= # moles

V must be in liters (change if necessary)

1. What is the molarity of a 0.30 liter solution containing 0.50 moles of NaCl?

1.7M

2. Calculate the molarity of 0.289 moles of FeCl3 dissolved in 120 ml of solution?

2.41 M

3. If a 0.075 liter solution contains 0.0877 moles of CuCO4, what is the molarity?

1.2M

4. How many moles of NaCl are present in 600. ml a 1.55 M NaCl solution?

.930 moles

5. How many moles of H2SO4 are present in 1.63 liters of a 0.954 M solution?

1.56 molse

6. How many liters of solution are needed to make a 1.66 M solution containing 2.11 moles of KMnO4?

1.27 g

7. What volume of a 0.25 M solution can be made using 0.55 moles of Ca(OH)2?

2.2 L

For all of the problems below you will need to do a mole-mass conversion. Each problem will involve two steps.

8. What is the molarity in 650. ml of solution containing 63 grams of NaCl?

1.7 M

9. How many grams of Ca(OH)2 are needed to produce 500. ml of 1.66 M Ca(OH)2 solution?

61.5 g

10. What volume of a 0.88 M solution can be made using 130. grams of FeCl2?

1.2 L

Dilution Problems Worksheet

1. How do you prepare a 250.-ml of a 2.35 M HF dilution from a 15.0 M stock solution?

39.2 mL

2. If 455-ml of 6.0 M HNO3 is used to make a 2.5 L dilution, what is the molarity of the dilution?

1.1 M

3. If 65.5 ml of HCl stock solution is used to make 450.-ml of a 0.675 M HCl dilution, what is the molarity of the stock solution?

4.64 M

4. How do you prepare 500.-ml of a 1.77 M H2SO4 dilution from an 18.0 M H2SO4 stock solution?

Take 49.2-ml of 18.0 M H2SO4 stock solution and pour it into a 500-ml volumetric flask. Fill to the 500-ml line with distilled water to make 1.77M H2SO4 solution.

Extra Molarity Problems for Practice

1. How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?

3.75 M

2. How many moles of Sr(NO3)2 would be used in the preparation of 2.50 L of a 3.5 M solution?

8.75 M

3. What is the molarity of a 500-ml solution containing 249 g of KI?

3.00 M

4. How many grams of CaCl2 would be required to produce a 3.5 M solution with a volume of 2.0 L?

777 g

Explanation:

I've done this before

6 0

3 years ago