Mathematical Analysis and Data Trends Alıştırma Soruları — ACT — Sayfa 2

Soru 21Soru →

In a biochemistry laboratory experiment, a microfluidic device is calibrated to pump an enzyme solution into a reaction chamber at a constant rate of $4.8 \times 10^{-5}$ liters per minute ( $\text{L/min}$ ). What is this pump rate in microliters per second ( $\mu\text{L/s}$ )?

Cevabı ve açıklamayı göster

Cevap: 0.8

Cevap

The correct pump rate is 0.8 microliters per second.

To convert

4.8 \times 10^{-5}\text{ L/min}

to

\mu\text{L/s}

, first convert liters to microliters by multiplying by

10^6\ \mu\text{L/L}

, which gives

48\ \mu\text{L/min}

. Next, convert minutes to seconds by dividing by

60\text{ s/min}

to get

0.8\ \mu\text{L/s}

.

Adım Adım Çözüm

1

Convert the volume flow rate from liters per minute to microliters per minute.

48 \mu\text{L/min}

Since

1\text{ L} = 10^6\ \mu\text{L}

, multiplying

4.8 \times 10^{-5}\text{ L/min}

by

10^6\ \mu\text{L/L}

yields

48\ \mu\text{L/min}

.

2

Convert the flow rate from microliters per minute to microliters per second.

0.8 \mu\text{L/s}

Since

1\text{ minute} = 60\text{ seconds}

, dividing

48\ \mu\text{L/min}

by

60\text{ s/min}

yields

0.8\ \mu\text{L/s}

.

Anahtar Kavram

Scientific Notation and Unit Conversions

Soru 22Soru →

During a laboratory study on cellular respiration, a researcher measures the oxygen consumption rate of a single isolated mitochondrion. The mitochondrion consumes $3.5 \times 10^{-15}\text{ moles}$ of oxygen gas ( $\text{O}_2$ ) per second. To compare this with macro-scale metabolic rates, the researcher needs to express this value in picomoles ( $\text{pmol}$ ) per hour. Which of the following is closest to the oxygen consumption rate of the mitochondrion in $\text{pmol/hour}$ ?

9.72 \times 10^{-7}\text{ pmol/hour}

1.26 \times 10^{-2}\text{ pmol/hour}

1.26 \times 10^1\text{ pmol/hour}

1.26 \times 10^4\text{ pmol/hour}

Cevabı ve açıklamayı göster

Cevap: $1.26 \times 10^1\text{ pmol/hour}$

Cevap

1.26 \times 10^1\text{ pmol/hour}

The correct option is obtained by performing a two-step dimensional analysis. First, convert moles to picomoles using the relationship

1\text{ mole} = 10^{12}\text{ pmol}

. This gives

3.5 \times 10^{-15} \times 10^{12} = 3.5 \times 10^{-3}\text{ pmol/second}

. Second, convert seconds to hours by multiplying by the conversion factor of

3,600\text{ seconds per hour}

. This calculation yields

3.5 \times 10^{-3} \times 3,600 = 12.6\text{ pmol/hour}

. Expressing this in scientific notation gives

1.26 \times 10^1\text{ pmol/hour}

.

Adım Adım Çözüm

1

Convert the oxygen consumption rate from moles per second to picomoles per second.

3.5 \times 10^{-15}\text{ moles/second} \times 10^{12}\text{ pmol/mole} = 3.5 \times 10^{-3}\text{ pmol/second}

Since

1\text{ picomole (pmol)} = 10^{-12}\text{ moles}

, there are

10^{12}\text{ picomoles}

in

1\text{ mole}

. Multiplying by this factor converts the unit of quantity.

2

Convert the rate from per second to per hour.

3.5 \times 10^{-3}\text{ pmol/second} \times 3,600\text{ seconds/hour} = 1.26 \times 10^1\text{ pmol/hour}

There are

3,600\text{ seconds}

in one hour. Multiplying the per-second rate by

3,600

gives the total consumption over the duration of an hour.

Anahtar Kavram

Scientific Notation and Unit Conversions

Soru 23Soru →

A student measured the speed of sound in a chamber filled with pure carbon dioxide ( $CO_2$ ) gas at various temperatures. The measured speed of sound, in meters per second ( $\text{m/s}$ ), at each temperature, in degrees Celsius ( $^\circ\text{C}$ ), is shown in the table below:

Temperature ( $^\circ\text{C}$ )	Speed of Sound ( $\text{m/s}$ )
$0$	$259$
$20$	$268$
$40$	$277$
$60$	$286$

Assuming the speed of sound continues to change at a constant rate with respect to temperature, what is the predicted speed of sound in $CO_2$ gas, in meters per second ( $\text{m/s}$ ), at a temperature of $100^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 304

Cevap

The predicted speed of sound in carbon dioxide gas at

100^\circ\text{C}

is

304\text{ m/s}

.

The speed of sound increases linearly by

9\text{ m/s}

for every

20^\circ\text{C}

increase in temperature, which is a rate of

0.45\text{ m/s}

per

1^\circ\text{C}

. The target temperature of

100^\circ\text{C}

is

40^\circ\text{C}

higher than the highest data point in the table (

60^\circ\text{C}

). The speed of sound will therefore increase by

40 \times 0.45 = 18\text{ m/s}

beyond the

60^\circ\text{C}

speed. Adding this to

286\text{ m/s}

yields

304\text{ m/s}

.

Adım Adım Çözüm

1

Determine the constant rate of change of the speed of sound per

1^\circ\text{C}

temperature increase.

The speed of sound increases at a rate of

0.45\text{ m/s}

per

1^\circ\text{C}

.

This establishes the linear trend shown in the experimental data.

2

Find the temperature interval between the highest measured data point and the target temperature.

The difference is

40^\circ\text{C}

(from

60^\circ\text{C}

to

100^\circ\text{C}

).

This determines how far outside the measured data range the extrapolation must extend.

3

Multiply the temperature interval by the rate of change and add it to the speed of sound at the highest measured temperature.

286\text{ m/s} + (40^\circ\text{C} \times 0.45\text{ m/s/}^\circ\text{C}) = 304\text{ m/s}

.

This completes the linear extrapolation to predict the final value.

Anahtar Kavram

Linear extrapolation relies on determining a constant rate of change from the given data points and applying it to a target value outside the experimental range.

Soru 24Soru →

A geophysicist studying volcanic emissions measures the mass of carbon dioxide ( $CO_2$ ) released by four different vents (Vents A, B, C, and D) in a hydrothermal field over a 1-hour period. The recorded masses are:

- Vent A: $4.2 \times 10^5 \text{ mg}$
- Vent B: $0.052 \text{ kg}$
- Vent C: $3.8 \times 10^3 \text{ cg}$
- Vent D: $8.9 \times 10^{-1} \text{ g}$

Based on these measurements, arrange the vents in order from the smallest mass of $CO_2$ released to the largest mass of $CO_2$ released.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct order of vents from smallest to largest mass of

CO_2

released is Vent D, Vent C, Vent B, then Vent A.

To compare the masses, they must be converted to a common unit, such as grams. Vent D releases

0.89 \text{ g}

, Vent C releases

38 \text{ g}

, Vent B releases

52 \text{ g}

, and Vent A releases

420 \text{ g}

. Comparing these quantities confirms the order from smallest to largest is Vent D, Vent C, Vent B, and Vent A.

Adım Adım Çözüm

1

Convert the mass of Vent A from milligrams to grams.

4.2 \times 10^5 \text{ mg} \times \frac{1 \text{ g}}{10^3 \text{ mg}} = 4.2 \times 10^2 \text{ g} = 420 \text{ g}

Converting all masses to a single standard unit (grams) allows for a direct comparison.

2

Convert the mass of Vent B from kilograms to grams.

0.052 \text{ kg} \times \frac{10^3 \text{ g}}{1 \text{ kg}} = 52 \text{ g}

Converting kilograms to grams requires multiplying by the conversion factor of

10^3 \text{ g/kg}

.

3

Convert the mass of Vent C from centigrams to grams.

3.8 \times 10^3 \text{ cg} \times \frac{1 \text{ g}}{10^2 \text{ cg}} = 3.8 \times 10^1 \text{ g} = 38 \text{ g}

Since centi- means

10^{-2}

, converting centigrams to grams requires dividing by

10^2

.

4

Convert the mass of Vent D to a standard decimal value in grams.

8.9 \times 10^{-1} \text{ g} = 0.89 \text{ g}

Expressing

8.9 \times 10^{-1} \text{ g}

in standard decimal format makes comparison straightforward.

5

Compare the converted masses in grams to order them from smallest to largest.

0.89 \text{ g} < 38 \text{ g} < 52 \text{ g} < 420 \text{ g}

Comparing the values shows that Vent D releases the least mass (

0.89 \text{ g}

), followed by Vent C (

38 \text{ g}

), Vent B (

52 \text{ g}

), and Vent A (

420 \text{ g}

).

Anahtar Kavram

Scientific Notation and Unit Conversions

Tahmini Süre:1m 30s

Soru 25Soru →

A group of researchers investigated the electrical properties of a newly developed conducting polymer. They performed two experiments to understand how the physical dimensions of a cylindrical polymer wire affect its electrical resistance, $R$ (in ohms, $\Omega$ ).

In Experiment 1, the researchers measured the resistance of polymer wires of various lengths, $L$ (in meters, $\text{m}$ ), while keeping the cross-sectional area constant at $A = 0.5\text{ mm}^2$ . The results are shown in Table 1.

Table 1

Trial	Wire Length $L$ ( $\text{m}$ )	Resistance $R$ ( $\Omega$ )
1	2.0	0.068
2	4.0	0.136
3	6.0	0.204
4	8.0	0.272

In Experiment 2, the researchers measured the resistance of polymer wires of a constant length $L = 5.0\text{ m}$ while varying the cross-sectional area, $A$ (in square millimeters, $\text{mm}^2$ ). The results are shown in Table 2.

Table 2

Trial	Cross-Sectional Area $A$ ( $\text{mm}^2$ )	Resistance $R$ ( $\Omega$ )
5	0.2	0.425
6	0.4	0.213
7	0.8	0.106
8	1.6	0.053

Based on the results of Experiments 1 and 2, what is the expected electrical resistance of a cylindrical polymer wire with a length of $12.0\text{ m}$ and a cross-sectional area of $0.3\text{ mm}^2$ ?

0.245\ \Omega

0.408\ \Omega

0.680\ \Omega

1.133\ \Omega

Cevabı ve açıklamayı göster

Cevap: $0.680\ \Omega$

Cevap

The expected electrical resistance of the polymer wire is

0.680\ \Omega

.

The correct answer is

0.680\ \Omega

. Experiment 1 shows that resistance

R

is directly proportional to length

L

because the ratio

R/L

is constant (

0.034\ \Omega/\text{m}

at

A = 0.5\text{ mm}^2

). Experiment 2 shows that resistance

R

is inversely proportional to area

A

because the product

R \times A

is constant (

0.085\ \Omega\cdot\text{mm}^2

at

L = 5.0\text{ m}

). Combining these gives

R = k \frac{L}{A}

. Solving for the constant gives

k = 0.017\ \Omega\cdot\text{mm}^2/\text{m}

. Applying this to a wire with

L = 12.0\text{ m}

and

A = 0.3\text{ mm}^2

yields

R = 0.017 \times \frac{12.0}{0.3} = 0.680\ \Omega

.

Adım Adım Çözüm

1

Determine the relationship between resistance (

R

) and wire length (

L

) from Experiment 1.

In Table 1, as the wire length

L

doubles (e.g., from

2.0\text{ m}

to

4.0\text{ m}

), the resistance

R

also doubles (from

0.068\ \Omega

to

0.136\ \Omega

). This indicates that resistance is directly proportional to wire length:

R \propto L

.

Establishing the relationship between length and resistance is necessary to scale the resistance for the new wire length.

2

Determine the relationship between resistance (

R

) and cross-sectional area (

A

) from Experiment 2.

In Table 2, as the area

A

doubles (e.g., from

0.2\text{ mm}^2

to

0.4\text{ mm}^2

), the resistance

R

is halved (from

0.425\ \Omega

to

0.213\ \Omega

). The product

R \times A

remains constant (

0.425 \times 0.2 \approx 0.085

). This indicates that resistance is inversely proportional to cross-sectional area:

R \propto \frac{1}{A}

.

Establishing the relationship between area and resistance is necessary to scale the resistance for the new cross-sectional area.

3

Combine the proportional relationships and find the constant of proportionality.

The combined relationship is

R = k \frac{L}{A}

, where

k

is a constant. Using Trial 1 where

L = 2.0\text{ m}

,

A = 0.5\text{ mm}^2

, and

R = 0.068\ \Omega

:

0.068 = k \frac{2.0}{0.5} \implies 0.068 = 4k \implies k = 0.017\ \Omega\cdot\text{mm}^2/\text{m}

.

Finding the formula and constant allows direct calculation of the resistance for any combination of length and area.

4

Calculate the resistance for a wire with

L = 12.0\text{ m}

and

A = 0.3\text{ mm}^2

.

R = 0.017 \times \frac{12.0}{0.3} = 0.017 \times 40 = 0.680\ \Omega

.

Applying the values to the combined formula yields the final expected resistance.

Anahtar Kavram

Direct and Inverse Proportionality

Alternatif Yöntem

Instead of finding the constant

k

, you can solve this using scaling factors. First, find the resistance for a wire of length

L_2 = 12.0\text{ m}

and the baseline area

A_1 = 0.5\text{ mm}^2

. Since

R

is directly proportional to

L

, scaling the length from

8.0\text{ m}

(Trial 4) to

12.0\text{ m}

scales the resistance by a factor of

\frac{12.0}{8.0} = 1.5

, giving

0.272 \times 1.5 = 0.408\ \Omega

. Next, adjust for the area change from

0.5\text{ mm}^2

to

0.3\text{ mm}^2

. Since

R

is inversely proportional to

A

, scaling the area by a factor of

\frac{0.3}{0.5} = 0.6

scales the resistance by a factor of

\frac{1}{0.6} = \frac{5}{3}

. This gives

0.408 \times \frac{5}{3} = 0.680\ \Omega

.

Tahmini Süre:1m 30s

Soru 26Soru →

A student conducted a series of trials to investigate Fick's first law of diffusion using a synthetic membrane. The rate of diffusion of a solute, $J$ (in milligrams per second, $\text{mg/s}$ ), is directly proportional to both the surface area of the membrane, $A$ (in square centimeters, $\text{cm}^2$ ), and the concentration difference of the solute across the membrane, $\Delta C$ (in moles per liter, $\text{mol/L}$ ), and is inversely proportional to the thickness of the membrane, $x$ (in millimeters, $\text{mm}$ ).

The parameters for Trial 1 and Trial 2 are shown in the table below:

Trial	Membrane thickness, $x$ ( $\text{mm}$ )	Membrane surface area, $A$ ( $\text{cm}^2$ )	Concentration difference, $\Delta C$ ( $\text{mol/L}$ )	Diffusion rate, $J$ ( $\text{mg/s}$ )
1	$0.20$	$3.0$	$0.06$	$27.0$
2	$0.60$	$8.0$	$0.04$	?

Based on the table, what was the resulting diffusion rate of the solute in Trial 2, in $\text{mg/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The diffusion rate of the solute in Trial 2 is 16.0 mg/s.

By setting up the proportionality equation

J = k \frac{A \cdot \Delta C}{x}

, we find the constant

k = 30

using the parameters from Trial 1. Substituting the parameters from Trial 2 yields a diffusion rate of exactly 16.0 mg/s.

Adım Adım Çözüm

1

Set up the algebraic relationship for the variables based on proportionality rules.

J = k \frac{A \cdot \Delta C}{x}

The rate of diffusion (

J

) is directly proportional to the surface area (

A

) and concentration difference (

\Delta C

), meaning they appear in the numerator. It is inversely proportional to membrane thickness (

x

), meaning it appears in the denominator. Here,

k

represents the constant of proportionality.

2

Calculate the constant of proportionality,

k

, using the values provided for Trial 1.

k = 30

Substitute

J_1 = 27.0

,

A_1 = 3.0

,

\Delta C_1 = 0.06

, and

x_1 = 0.20

into the equation:

27.0 = k \frac{3.0 \times 0.06}{0.20} \implies 27.0 = 0.9k \implies k = 30

.

3

Calculate the diffusion rate for Trial 2,

J_2

, by substituting the new parameters and the calculated constant

k

into the equation.

J_2 = 16.0

Substitute

k = 30

,

A_2 = 8.0

,

\Delta C_2 = 0.04

, and

x_2 = 0.60

into the equation:

J_2 = 30 \times \frac{8.0 \times 0.04}{0.60} = 30 \times \frac{0.32}{0.60} = 16.0

.

Anahtar Kavram

Direct and Inverse Proportionality

Soru 27Soru →

A student conducted a study on parallel-plate capacitors to determine how capacitance is affected by physical dimensions. In Experiment 1, the student set the distance between the plates to a constant value $d_1$ and measured the capacitance ( $C$ , in picofarads, $\text{pF}$ ) for plates of various surface areas ( $A$ , in $\text{cm}^2$ ). The results are shown in Table 1.

Plate Area $A$ ( $\text{cm}^2$ )	Capacitance $C$ ( $\text{pF}$ )
10.0	8.8
20.0	17.6
30.0	26.4
40.0	35.2

In Experiment 2, the student used plates of a constant surface area $A_1$ and measured the capacitance at various plate separation distances ( $d$ , in millimeters, $\text{mm}$ ). The results are shown in Table 2.

Plate Separation $d$ ( $\text{mm}$ )	Capacitance $C$ ( $\text{pF}$ )
1.0	35.2
2.0	17.6
4.0	8.8
8.0	4.4

Based on these results, if the student constructs a capacitor using the same materials with a plate area of $60.0\text{ cm}^2$ and a plate separation distance of $4.0\text{ mm}$ , what will be the expected capacitance of this capacitor?

5.9\text{ pF}

13.2\text{ pF}

93.9\text{ pF}

211.2\text{ pF}

Cevabı ve açıklamayı göster

Cevap: $13.2\text{ pF}$

Cevap

The expected capacitance of the capacitor is

13.2\text{ pF}

.

The data shows that capacitance is directly proportional to plate area and inversely proportional to plate separation distance. Starting from a baseline of

A = 40.0\text{ cm}^2

and

d = 1.0\text{ mm}

(where

C = 35.2\text{ pF}

), increasing the area to

60.0\text{ cm}^2

(a factor of

1.5

) increases the capacitance to

52.8\text{ pF}

. Then, increasing the separation distance to

4.0\text{ mm}

(a factor of

4

) divides the capacitance by

4

, resulting in

13.2\text{ pF}

.

Adım Adım Çözüm

1

Determine the proportional relationship between capacitance and plate area from Experiment 1.

Capacitance (

C

) is directly proportional to plate area (

A

) because doubling the area (e.g., from

10.0\text{ cm}^2

to

20.0\text{ cm}^2

) doubles the capacitance (from

8.8\text{ pF}

to

17.6\text{ pF}

), meaning

C \propto A

.

Establishing the relationship for the first independent variable is necessary to scale its effect.

2

Determine the proportional relationship between capacitance and plate separation distance from Experiment 2.

Capacitance (

C

) is inversely proportional to plate separation (

d

) because doubling the distance (e.g., from

1.0\text{ mm}

to

2.0\text{ mm}

) halves the capacitance (from

35.2\text{ pF}

to

17.6\text{ pF}

), meaning

C \propto \frac{1}{d}

.

Establishing the relationship for the second independent variable is necessary to scale its effect.

3

Select a baseline configuration from the data to perform the scaling calculation.

Using the configuration from Table 2 where

d = 1.0\text{ mm}

and

C = 35.2\text{ pF}

. From Table 1, we see this corresponds to a constant plate area

A_1 = 40.0\text{ cm}^2

.

A known reference point with both dimensions and capacitance is needed as a starting point.

4

Scale the baseline capacitance for the change in plate area from

40.0\text{ cm}^2

to

60.0\text{ cm}^2

.

The area increases by a factor of

\frac{60.0}{40.0} = 1.5

. Since

C

is directly proportional to

A

, the capacitance increases to

35.2\text{ pF} \times 1.5 = 52.8\text{ pF}

at

d = 1.0\text{ mm}

.

This accounts for the direct scaling of the plate area.

5

Scale the intermediate capacitance for the change in plate separation from

1.0\text{ mm}

to

4.0\text{ mm}

.

The separation distance increases by a factor of

\frac{4.0}{1.0} = 4

. Since

C

is inversely proportional to

d

, the capacitance is divided by

4

, yielding

\frac{52.8\text{ pF}}{4} = 13.2\text{ pF}

.

This accounts for the inverse scaling of the plate separation distance.

Anahtar Kavram

Direct and Inverse Proportionality in Experimental Data

Soru 28Soru →

### Passage

A student conducts three experiments to study the fundamental frequency, $f$ (in hertz, $\text{Hz}$ ), of a vibrating string on a sonometer.

In Experiment 1, the student varies the length of the string, $L$ (in meters, $\text{m}$ ), while keeping the tension, $T$ (in newtons, $\text{N}$ ), and the linear mass density, $\mu$ (in grams per meter, $\text{g/m}$ ), constant.

In Experiment 2, the student varies the tension, $T$ , while keeping the length ( $L = 0.50\ \text{m}$ ) and linear mass density ( $\mu = 2.0\ \text{g/m}$ ) constant.

In Experiment 3, the student varies the linear mass density, $\mu$ , by using different strings while keeping the length ( $L = 0.50\ \text{m}$ ) and tension ( $T = 100\ \text{N}$ ) constant.

The results of the three experiments are recorded in the tables below:

Table 1 (Experiment 1)

Trial	Length $L$ ( $\text{m}$ )	Frequency $f$ ( $\text{Hz}$ )
1	$0.25$	$440$
2	$0.50$	$220$
3	$1.00$	$110$

Table 2 (Experiment 2)

Trial	Tension $T$ ( $\text{N}$ )	Frequency $f$ ( $\text{Hz}$ )
4	$25$	$110$
5	$100$	$220$
6	$400$	$440$

Table 3 (Experiment 3)

Trial	Linear mass density $\mu$ ( $\text{g/m}$ )	Frequency $f$ ( $\text{Hz}$ )
7	$0.5$	$440$
8	$2.0$	$220$
9	$8.0$	$110$

Based on the tables, match each physical relationship to the equation that correctly describes the proportionality and fits the experimental data.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Frequency as a function of length is

f = \frac{110}{L}

; frequency as a function of tension is

f = 22\sqrt{T}

; the square of frequency as a function of linear mass density is

f^2 = \frac{96,800}{\mu}

; and frequency as a function of linear mass density is

f = \frac{220\sqrt{2}}{\sqrt{\mu}}

.

The correct pairings are established by identifying the constant product or ratio for each set of experimental trials. For Experiment 1, the product

f \cdot L

is constant at

110

, showing an inverse relationship

f = \frac{110}{L}

. For Experiment 2, the ratio

\frac{f}{\sqrt{T}}

is constant at

22

, showing a direct relationship to the square root,

f = 22\sqrt{T}

. For Experiment 3, the product

f^2 \cdot \mu

is constant at

96,800

, showing that

f^2

is inversely proportional to

\mu

, which simplifies to

f = \frac{220\sqrt{2}}{\sqrt{\mu}}

.

Adım Adım Çözüm

1

Analyze the relationship between frequency (

f

) and string length (

L

) using Table 1.

Doubling

L

from

0.25\ \text{m}

to

0.50\ \text{m}

halves

f

from

440\ \text{Hz}

to

220\ \text{Hz}

, indicating that

f

is inversely proportional to

L

(

f = \frac{k_1}{L}

). Solving for the constant gives

k_1 = f \cdot L = 220 \times 0.50 = 110

. Thus,

f = \frac{110}{L}

.

To determine the equation for frequency as a function of length under constant tension and linear mass density.

2

Analyze the relationship between frequency (

f

) and tension (

T

) using Table 2.

Quadrupling

T

from

25\ \text{N}

to

100\ \text{N}

doubles

f

from

110\ \text{Hz}

to

220\ \text{Hz}

, indicating that

f

is directly proportional to the square root of tension (

f = k_2\sqrt{T}

). Solving for the constant gives

k_2 = \frac{f}{\sqrt{T}} = \frac{220}{\sqrt{100}} = 22

. Thus,

f = 22\sqrt{T}

.

To determine the equation for frequency as a function of tension under constant length and linear mass density.

3

Analyze the relationship between the square of the frequency (

f^2

) and linear mass density (

\mu

) using Table 3.

Quadrupling

\mu

from

0.5\ \text{g/m}

to

2.0\ \text{g/m}

halves

f

from

440\ \text{Hz}

to

220\ \text{Hz}

, meaning that

f^2

is quartered from

193,600\ \text{Hz}^2

to

48,400\ \text{Hz}^2

. This shows that

f^2

is inversely proportional to

\mu

(

f^2 = \frac{k_3}{\mu}

). Solving for the constant gives

k_3 = f^2 \cdot \mu = 220^2 \times 2.0 = 96,800

. Thus,

f^2 = \frac{96,800}{\mu}

.

To determine the equation for the square of the frequency as a function of linear mass density.

4

Derive the direct relationship between frequency (

f

) and linear mass density (

\mu

) using the equation from Step 3.

Taking the square root of

f^2 = \frac{96,800}{\mu}

yields

f = \frac{\sqrt{96,800}}{\sqrt{\mu}} = \frac{220\sqrt{2}}{\sqrt{\mu}}

.

To express the frequency as a function of the square root of linear mass density.

Anahtar Kavram

Direct and inverse proportionality in physical systems, including relationships involving roots and powers of variables.

Tahmini Süre:2m 0s

Soru 29Soru →

Aerodynamic drag force ( $F_d$ ) acts on vehicles as they move through the air. A group of students measured the drag force, in newtons ( $\text{N}$ ), acting on a scale model of a sports car in a wind tunnel at various wind velocities ( $v$ ), in meters per second ( $\text{m/s}$ ). The data from their trials are recorded in the table below:

Velocity ( $v$ , $\text{m/s}$ )	Drag Force ( $F_d$ , $\text{N}$ )
$10$	$12$
$20$	$48$
$30$	$108$
$40$	$192$

Based on the trend shown in the table, what is the expected aerodynamic drag force, in newtons ( $\text{N}$ ), acting on the scale model when the wind velocity is $50\text{ m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 300

Cevap

The expected aerodynamic drag force is 300 N.

The correct answer is 300 N because the drag force scales quadratically with velocity according to the relation

F_d = 0.12 v^2

. Plugging in

v = 50\text{ m/s}

yields

0.12 \times 2500 = 300\text{ N}

.

Adım Adım Çözüm

1

Calculate the ratio of drag force to the square of the velocity for the given data points.

For all data points,

F_d / v^2 = 0.12

. This establishes the quadratic trend

F_d = 0.12 v^2

.

Identifying the mathematical relationship between the variables is necessary to accurately extrapolate beyond the measured data range.

2

Substitute the target velocity of

50\text{ m/s}

into the identified quadratic formula.

F_d = 0.12 \times (50)^2 = 300\text{ N}

.

Applying the mathematical trend allows for the calculation of the drag force at the extrapolated velocity.

Anahtar Kavram

Extrapolation of a quadratic relationship between velocity and aerodynamic drag force.

Tahmini Süre:1m 30s

Soru 30Soru →

Geologists drilled a deep borehole into the Earth's crust at a research site and measured the rock temperature at various depths. The recorded temperatures are shown in the table below:

Depth ( $m$ )	Temperature ( $^\circ\text{C}$ )
$0$	$12.0$
$250$	$19.5$
$500$	$27.0$
$750$	$34.5$
$1,000$	$42.0$

Assuming the temperature continues to increase linearly with depth at the same rate observed between $0\text{ m}$ and $1,000\text{ m}$ , what will the rock temperature, in degrees Celsius ( $^\circ\text{C}$ ), be at a depth of $1,800\text{ m}$ ?

Cevabı ve açıklamayı göster

Cevap: 66

Cevap

The projected rock temperature at a depth of 1,800 m is 66.0°C.

The correct calculation determines that the temperature increases by 7.5°C for every 250 m (a rate of 0.03°C/m). Extrapolating linearly to 1,800 m, the temperature increases by 54.0°C (0.03°C/m * 1,800 m) from the surface baseline of 12.0°C, yielding a final temperature of 66.0°C.

Adım Adım Çözüm

1

Calculate the rate of temperature change per meter of depth.

0.03^\circ\text{C/m}

Using the interval from

0\text{ m}

to

250\text{ m}

, the temperature increases by

19.5^\circ\text{C} - 12.0^\circ\text{C} = 7.5^\circ\text{C}

. The rate of change is

\frac{7.5^\circ\text{C}}{250\text{ m}} = 0.03^\circ\text{C/m}

.

2

Calculate the total temperature change over the depth interval of 1,800 m.

54.0^\circ\text{C}

Multiplying the constant rate of temperature change (

0.03^\circ\text{C/m}

) by the target depth (

1,800\text{ m}

) yields the total increase in temperature from the surface:

0.03 \times 1,800 = 54.0^\circ\text{C}

.

3

Determine the final temperature by adding the increase to the baseline surface temperature.

66.0^\circ\text{C}

Adding the

54.0^\circ\text{C}

increase to the baseline temperature at the surface (

12.0^\circ\text{C}

) gives the projected temperature at depth:

12.0^\circ\text{C} + 54.0^\circ\text{C} = 66.0^\circ\text{C}

.

Anahtar Kavram

Extrapolation of linear data trends

Tahmini Süre:1m 30s

Soru 31Soru →

A student conducted two experiments to investigate the magnetic field strength, $B$ (measured in microtesla, $\mu\text{T}$ ), surrounding a straight, current-carrying wire. In Experiment 1, the student measured the magnetic field at varying distances, $r$ , from the wire while holding the current constant. In Experiment 2, the student measured the magnetic field at a fixed distance while varying the current, $I$ . The results of these experiments are shown in Table 1 and Table 2.

### Table 1

Trial	Current $I$ ( $\text{A}$ )	Distance $r$ ( $\text{m}$ )	Magnetic Field $B$ ( $\mu\text{T}$ )
1	2.0	0.10	4.0
2	2.0	0.20	2.0
3	2.0	0.40	1.0

### Table 2

Trial	Distance $r$ ( $\text{m}$ )	Current $I$ ( $\text{A}$ )	Magnetic Field $B$ ( $\mu\text{T}$ )
4	0.10	1.0	2.0
5	0.10	2.0	4.0
6	0.10	3.0	6.0

Based on the results of the experiments, if the student conducts a new trial with a current of $4.0\text{ A}$ at a distance of $0.20\text{ m}$ from the wire, what is the predicted magnetic field strength $B$ ?

1.0\ \mu\text{T}

2.0\ \mu\text{T}

4.0\ \mu\text{T}

16.0\ \mu\text{T}

Cevabı ve açıklamayı göster

Cevap: $4.0\ \mu\text{T}$

Cevap

4.0\ \mu\text{T}

The correct answer is

4.0\ \mu\text{T}

. According to Table 1, when current is held constant, doubling the distance from

0.10\text{ m}

to

0.20\text{ m}

halves the magnetic field strength from

4.0\ \mu\text{T}

to

2.0\ \mu\text{T}

, demonstrating an inverse proportionality. According to Table 2, when distance is held constant, doubling the current from

1.0\text{ A}

to

2.0\text{ A}

doubles the magnetic field strength from

2.0\ \mu\text{T}

to

4.0\ \mu\text{T}

, demonstrating a direct proportionality. Starting from Trial 2 where

r = 0.20\text{ m}

,

I = 2.0\text{ A}

, and

B = 2.0\ \mu\text{T}

, doubling the current to

4.0\text{ A}

while keeping the distance constant at

0.20\text{ m}

doubles the field strength to

4.0\ \mu\text{T}

.

Adım Adım Çözüm

1

Determine the relationship between distance

r

and magnetic field

B

.

B

is inversely proportional to

r

(

B \propto 1/r

).

According to Table 1, with current constant, doubling the distance from

0.10\text{ m}

to

0.20\text{ m}

reduces the magnetic field from

4.0\ \mu\text{T}

to

2.0\ \mu\text{T}

.

2

Determine the relationship between current

I

and magnetic field

B

.

B

is directly proportional to

I

(

B \propto I

).

According to Table 2, with distance constant, doubling the current from

1.0\text{ A}

to

2.0\text{ A}

doubles the magnetic field from

2.0\ \mu\text{T}

to

4.0\ \mu\text{T}

.

3

Calculate the predicted magnetic field for a new trial with

I = 4.0\text{ A}

and

r = 0.20\text{ m}

using a baseline trial.

B = 4.0\ \mu\text{T}

Starting from Trial 2 (

I = 2.0\text{ A}

,

r = 0.20\text{ m}

,

B = 2.0\ \mu\text{T}

), the distance matches the target trial. Since current is doubled from

2.0\text{ A}

to

4.0\text{ A}

, and

B

is directly proportional to

I

, we multiply the field by

2

:

2.0\ \mu\text{T} \times 2 = 4.0\ \mu\text{T}

.

Anahtar Kavram

Direct and Inverse Proportionality

Tahmini Süre:1m 30s

Soru 32Soru →

A student conducted a series of trials using a rotating mass apparatus to investigate the relationship between centripetal force ( $F_c$ ) and the radius of rotation ( $r$ ) for a constant mass moving at a constant speed. The results of the trials are shown in the table below:

Trial	Radius ( $r$ , $\text{m}$ )	Centripetal Force ( $F_c$ , $\text{N}$ )
1	$0.80$	$12.0$
2	$2.40$	?

Given that the centripetal force is inversely proportional to the radius of rotation under these conditions, what is the centripetal force, in newtons, for Trial 2?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The centripetal force in Trial 2 is

4.0\text{ N}

.

Since centripetal force is inversely proportional to the radius of rotation, their product must remain constant (

F_c \times r = k

). Using Trial 1, we find

k = 12.0 \times 0.80 = 9.6

. For Trial 2, we set up the equation

F_c \times 2.40 = 9.6

, which yields

F_c = 4.0\text{ N}

.

Adım Adım Çözüm

1

Identify the mathematical relationship between the variables.

F_c \times r = k

The problem states that centripetal force is inversely proportional to the radius of rotation.

2

Calculate the constant of proportionality (

k

) using the data from Trial 1.

k = 12.0 \times 0.80 = 9.6

Both the radius (

r = 0.80\text{ m}

) and centripetal force (

F_c = 12.0\text{ N}

) are known for Trial 1.

3

Calculate the unknown centripetal force for Trial 2.

F_c = \frac{9.6}{2.40} = 4.0

The constant of proportionality is

9.6

and the radius for Trial 2 is

2.40\text{ m}

.

Anahtar Kavram

Inverse Proportionality

Soru 33Soru →

To study the properties of electrical conductors, a researcher measured the electrical resistance, $R$ (in ohms, $\Omega$ ), of four copper wires. The wires all have the same length but differ in their cross-sectional area, $A$ (in square millimeters, $\text{mm}^2$ ). The measurements are presented in the table below:

Wire	Cross-sectional area, $A$ ( $\text{mm}^2$ )	Resistance, $R$ ( $\Omega$ )
1	0.50	3.44
2	1.00	1.72
3	2.00	0.86
4	4.00	0.43

Based on the table, if a fifth wire of the same length and material has a cross-sectional area of $8.00\text{ mm}^2$ , what is its predicted resistance?

0.215\ \Omega

0.860\ \Omega

0.000\ \Omega

3.440\ \Omega

Cevabı ve açıklamayı göster

Cevap: $0.215\ \Omega$

Cevap

0.215\ \Omega

The correct answer is

0.215\ \Omega

. The data shows that the electrical resistance (

R

) is inversely proportional to the cross-sectional area (

A

) of the wire, because their product remains constant (

A \times R = 1.72

). Therefore, doubling the cross-sectional area from

4.00\text{ mm}^2

to

8.00\text{ mm}^2

requires halving the resistance from

0.43\ \Omega

to

0.215\ \Omega

.

Adım Adım Çözüm

1

Determine the mathematical relationship between cross-sectional area (

A

) and resistance (

R

) from the table.

The product of

A

and

R

is constant for all trials (

0.50 \times 3.44 = 1.72

,

1.00 \times 1.72 = 1.72

,

2.00 \times 0.86 = 1.72

, and

4.00 \times 0.43 = 1.72

). This demonstrates that

R

is inversely proportional to

A

, with the relationship

R = 1.72 / A

.

Identifying whether the relationship is direct or inverse allows us to correctly scale the variables.

2

Calculate the predicted resistance (

R

) for a wire with a cross-sectional area of

8.00\text{ mm}^2

.

R = 1.72 / 8.00 = 0.215\ \Omega

. Alternatively, since the area doubles from

4.00\text{ mm}^2

to

8.00\text{ mm}^2

, the resistance must be halved:

0.43\ \Omega / 2 = 0.215\ \Omega

.

Applying the constant of proportionality or the scaling factor determines the final value.

Anahtar Kavram

Direct and Inverse Proportionality

Soru 34Soru →

A student conducts three trials to investigate the mathematical relationships between voltage ( $V$ ), current ( $I$ ), resistance ( $R$ ), and electric power ( $P$ ) in a DC circuit. The data collected from these trials are shown in the tables below:

**Trial 1 (Constant Resistance of

10\ \Omega

)**

Voltage ( $V$ , $\text{V}$ )	Current ( $I$ , $\text{A}$ )
$2.0$	$0.20$
$4.0$	$0.40$
$6.0$	$0.60$

**Trial 2 (Constant Voltage of

12\ \text{V}

)**

Resistance ( $R$ , $\Omega$ )	Current ( $I$ , $\text{A}$ )
$2.0$	$6.0$
$4.0$	$3.0$
$6.0$	$2.0$

**Trial 3 (Constant Resistance of

2.0\ \Omega

)**

Current ( $I$ , $\text{A}$ )	Power ( $P$ , $\text{W}$ )
$1.0$	$2.0$
$2.0$	$8.0$
$3.0$	$18.0$

Based on the tables, match each trial to the mathematical relationship that best describes the variables in that trial.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Trial 1 matches direct linear proportionality; Trial 2 matches inverse proportionality; Trial 3 matches direct quadratic proportionality.

The correct matches align each trial with its proportional trend: Trial 1 displays a constant ratio between current and voltage, indicating direct linear proportionality. Trial 2 shows a constant product between current and resistance, indicating inverse proportionality. Trial 3 shows that power increases with the square of the current, indicating direct quadratic proportionality.

Adım Adım Çözüm

1

Analyze Trial 1 to determine the relationship between Voltage (

V

) and Current (

I

).

As voltage increases, current increases at a constant rate. Specifically, doubling voltage from

2.0\ \text{V}

to

4.0\ \text{V}

doubles the current from

0.20\ \text{A}

to

0.40\ \text{A}

. The constant ratio

\frac{I}{V} = 0.10\ \text{A/V}

confirms direct linear proportionality (

I \propto V

).

This determines the constant of proportionality and the nature of the relationship when both variables change in the same direction at a constant ratio.

2

Analyze Trial 2 to determine the relationship between Resistance (

R

) and Current (

I

).

As resistance increases, current decreases. Doubling the resistance from

2.0\ \Omega

to

4.0\ \Omega

halves the current from

6.0\ \text{A}

to

3.0\ \text{A}

. The product

I \times R = 12.0

remains constant, confirming inverse proportionality (

I \propto \frac{1}{R}

).

This identifies whether the variables have a constant product, which is the defining characteristic of an inverse relationship.

3

Analyze Trial 3 to determine the relationship between Current (

I

) and Power (

P

).

As current increases, power increases non-linearly. When current doubles from

1.0\ \text{A}

to

2.0\ \text{A}

, power increases by a factor of

4

(

2.0\ \text{W}

to

8.0\ \text{W}

). When current triples from

1.0\ \text{A}

to

3.0\ \text{A}

, power increases by a factor of

9

(

2.0\ \text{W}

to

18.0\ \text{W}

). This is a quadratic relationship, representing direct quadratic proportionality (

P \propto I^2

).

This distinguishes a linear increase from an exponential or power-based increase by calculating the factor changes.

Anahtar Kavram

Identifying direct linear, inverse, and quadratic proportional relationships from experimental tables by analyzing how proportional changes in the independent variable affect the dependent variable.

Tahmini Süre:1m 30s

Soru 35Soru →

A group of students conducted an experiment to study the relationship between the pressure and volume of a gas sample at a constant temperature. The gas was contained in a cylinder equipped with a movable piston. By adjusting the piston, the students varied the volume of the gas ( $V$ , in liters) and measured the resulting pressure ( $P$ , in atmospheres). The results are recorded in the table below.

Trial	Volume ( $V$ , $\text{L}$ )	Pressure ( $P$ , $\text{atm}$ )
1	$1.5$	$8.0$
2	$3.0$	$4.0$
3	$6.0$	$2.0$
4	$12.0$	$1.0$

If the students perform a fifth trial and adjust the piston to a volume of $4.0\text{ L}$ , which of the following is the most likely pressure of the gas sample?

1.3\text{ atm}

3.0\text{ atm}

5.3\text{ atm}

48.0\text{ atm}

Cevabı ve açıklamayı göster

Cevap: $3.0\text{ atm}$

Cevap

3.0\text{ atm}

The correct answer is the option stating

3.0\text{ atm}

. The data in the table shows that pressure (

P

) and volume (

V

) are inversely proportional because their product is constant across all trials:

P \times V = 1.5 \times 8.0 = 3.0 \times 4.0 = 6.0 \times 2.0 = 12.0\text{ L}\cdot\text{atm}

. For a volume of

4.0\text{ L}

, the pressure is calculated as

P = \frac{12.0}{4.0} = 3.0\text{ atm}

.

Adım Adım Çözüm

1

Analyze the relationship between volume (

V

) and pressure (

P

) in the table.

As the volume increases from

1.5

to

12.0\text{ L}

, the pressure decreases from

8.0

to

1.0\text{ atm}

. The product of pressure and volume remains constant for all trials:

P \times V = 12.0\text{ L}\cdot\text{atm}

. This indicates that pressure and volume are inversely proportional.

To identify whether the variables have a direct or inverse relationship and determine the constant of proportionality.

2

Apply the inverse proportionality equation to find the pressure for the new volume of

4.0\text{ L}

.

P = \frac{12.0\text{ L}\cdot\text{atm}}{4.0\text{ L}} = 3.0\text{ atm}

.

To calculate the expected pressure when the volume is set to

4.0\text{ L}

using the determined relationship.

Anahtar Kavram

Inverse proportionality dictates that as one variable increases, the other decreases such that their product remains constant (

y \propto \frac{1}{x}

, or

x \times y = k

).

Tahmini Süre:1m 30s

Soru 36Soru →

A student set up an experiment to investigate the mechanical properties of a simple two-gear system consisting of a driver gear and a driven gear. The driver gear rotates at a constant speed, while various driven gears with different numbers of teeth, $N$ , are swapped into the system. The student measured the rotational speed, $S$ (in revolutions per minute, rpm), of each driven gear. The table below shows the results of the experiment:

Driven Gear	Number of Teeth ( $N$ )	Rotational Speed ( $S$ , rpm)
Gear 1	8	270
Gear 2	12	180
Gear 3	16	135
Gear 4	18	?

Given that the rotational speed of the driven gear is inversely proportional to its number of teeth, what is the rotational speed, in rpm, of Gear 4?

Cevabı ve açıklamayı göster

Cevap: 120

Cevap

The rotational speed of Gear 4 is 120 rpm.

Since rotational speed is inversely proportional to the number of teeth, the product of these two values must remain constant. For all measured gears,

S \times N = 2160

. Dividing this constant by the 18 teeth of Gear 4 yields a rotational speed of 120 rpm.

Adım Adım Çözüm

1

Determine the proportionality constant

Constant

k = 2160

Because rotational speed (

S

) and number of teeth (

N

) are inversely proportional, their product is constant (

S \times N = k

). Using Gear 2 data,

12 \times 180 = 2160

.

2

Calculate the speed for Gear 4

Rotational speed = 120

Using the constant

k = 2160

and the number of teeth for Gear 4 (

N = 18

), the speed is calculated as

S = \frac{2160}{18} = 120

.

Anahtar Kavram

Direct and Inverse Proportionality

Soru 37Soru →

A group of students conducted a series of trials to study the interference patterns of light using a double-slit setup. They measured the fringe spacing, $w$ (the distance between adjacent bright bands on a screen). The relationship between the fringe spacing and the experimental parameters is given by:

w = \frac{\lambda L}{d}

where $\lambda$ is the wavelength of the light source, $L$ is the distance from the slits to the screen, and $d$ is the distance between the two slits.

Match each change in the experimental setup to its corresponding effect on the fringe spacing ( $w$ ).

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

To solve this, match the physical changes to their mathematical consequences: doubling the slit separation (

d

) halves the fringe spacing (

w

); doubling the wavelength (

\lambda

) doubles the fringe spacing (

w

); and doubling both the slit separation (

d

) and the distance to the screen (

L

) keeps the fringe spacing (

w

) unchanged.

The correct matches are based on the algebraic relationship

w = \frac{\lambda L}{d}

. Doubling a variable in the numerator (

\lambda

) doubles the value of

w

due to direct proportionality. Doubling a variable in the denominator (

d

) halves the value of

w

due to inverse proportionality. Doubling both simultaneously cancels the changes out (

\frac{2}{2} = 1

), keeping

w

constant.

Adım Adım Çözüm

1

Identify the proportional relationships for each variable in the equation

w = \frac{\lambda L}{d}

.

w

is directly proportional to

\lambda

and

L

, and inversely proportional to

d

.

This allows us to determine how changing each variable independently affects the value of

w

.

2

Analyze the effect of doubling the slit separation (

d

).

w' = \frac{\lambda L}{2d} = \frac{1}{2}w

.

Because

w

is inversely proportional to

d

, doubling

d

must result in halving

w

.

3

Analyze the effect of doubling the wavelength (

\lambda

).

w' = \frac{2\lambda L}{d} = 2w

.

Because

w

is directly proportional to

\lambda

, doubling

\lambda

must result in doubling

w

.

4

Analyze the effect of simultaneously doubling the slit separation (

d

) and the screen distance (

L

).

w' = \frac{\lambda (2L)}{2d} = \frac{2}{2}w = w

.

The direct proportionality factor of 2 from

L

and the inverse proportionality factor of 2 from

d

cancel each other out, leaving the fringe spacing unchanged.

Anahtar Kavram

Direct and inverse proportionality in algebraic equations

Tahmini Süre:1m 30s

Soru 38Soru →

A group of students conducted a physics experiment to study the relationship between the mass of a glider and its acceleration on a horizontal air track. A constant net force was applied to the glider during all trials. The students observed that the acceleration of the glider, $a$ , is inversely proportional to its mass, $m$ . During Trial 1, a glider with a mass of $0.20\text{ kg}$ was measured to have an acceleration of $5.0\text{ m/s}^2$ . During Trial 2, a different glider was used under the same constant net force. If the mass of the glider in Trial 2 is $0.50\text{ kg}$ , what is the acceleration of the glider, in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The acceleration of the glider in Trial 2 is 2.0 m/s^2.

Since acceleration is inversely proportional to mass, their product remains constant under a constant net force. Using the data from Trial 1, the constant is calculated as

0.20\text{ kg} \times 5.0\text{ m/s}^2 = 1.0\text{ N}

. In Trial 2, with a mass of

0.50\text{ kg}

, the acceleration is found by dividing the constant by the new mass:

1.0 / 0.50 = 2.0\text{ m/s}^2

.

Adım Adım Çözüm

1

Set up the inverse proportionality equation.

a \times m = k

Since acceleration is inversely proportional to mass, their product must equal a constant value.

2

Calculate the constant of proportionality using Trial 1 data.

k = 1.0

Substitute

m = 0.20\text{ kg}

and

a = 5.0\text{ m/s}^2

into the equation:

5.0 \times 0.20 = 1.0

.

3

Calculate the new acceleration for Trial 2.

a_2 = 2.0\text{ m/s}^2

Substitute the constant

k = 1.0

and the new mass

m_2 = 0.50\text{ kg}

into the equation:

a_2 \times 0.50 = 1.0

, so

a_2 = 2.0

.

Anahtar Kavram

Inverse proportionality relates two variables such that their product is constant. If one variable increases by a factor, the other must decrease by the same factor.

Soru 39Soru →

A researcher measured the flow rate, $Q$ , of various liquids passing through a narrow capillary tube under constant pressure. The viscosity, $\eta$ , of each liquid and its corresponding flow rate are shown in the table.

Liquid	Viscosity ( $\eta$ , $\text{mPa}\cdot\text{s}$ )	Flow rate ( $Q$ , $\text{mL/s}$ )
Liquid 1	$2.0$	$18.0$
Liquid 2	$3.0$	$12.0$
Liquid 3	$4.0$	$9.0$
Liquid 4	$6.0$	$6.0$

Based on these results, which of the following equations best represents the relationship between the flow rate, $Q$ , and the viscosity, $\eta$ , of the liquids?

Q = 36.0\eta

Q = \frac{\eta}{36.0}

Q = \frac{36.0}{\eta}

Q = 36.0 - \eta

Cevabı ve açıklamayı göster

Cevap: $Q = \frac{36.0}{\eta}$

Cevap

Q = \frac{36.0}{\eta}

The correct equation is the one stating that flow rate is equal to

36.0

divided by viscosity. In an inverse proportionality relationship, the product of the two variables is constant (

y \times x = k

). Multiplying the viscosity by the flow rate for each liquid in the table consistently yields

36.0

(e.g.,

2.0 \times 18.0 = 36.0

). Solving the equation

Q \times \eta = 36.0

for

Q

results in

Q = \frac{36.0}{\eta}

.

Adım Adım Çözüm

1

Analyze the trend in the data table between the viscosity,

\eta

, and the flow rate,

Q

.

As viscosity increases from

2.0

to

6.0

, the flow rate decreases from

18.0

to

6.0

. This indicates an inverse relationship between the two variables.

Identifying whether the relationship is direct or inverse helps eliminate incorrect equation forms.

2

Determine if the product of the two variables is constant, which is a characteristic of inverse proportionality (

y \cdot x = k

).

Calculate the product

Q \times \eta

for each trial:
- For Liquid 1:

2.0 \times 18.0 = 36.0

- For Liquid 2:

3.0 \times 12.0 = 36.0

- For Liquid 3:

4.0 \times 9.0 = 36.0

- For Liquid 4:

6.0 \times 6.0 = 36.0

The product is constant at

36.0

.

Finding the constant of proportionality establishes the exact mathematical relationship.

3

Formulate the equation expressing

Q

in terms of

\eta

.

Since

Q \times \eta = 36.0

, solving for

Q

gives

Q = \frac{36.0}{\eta}

.

This matches the target variable representation requested in the prompt.

Anahtar Kavram

Inverse Proportionality

Soru 40Soru →

A student conducted a series of trials to study how the volume of a gas sample changes as its temperature is varied under constant pressure. The data from the trials are shown in the table below:

Trial	Temperature ( $T$ , in $\text{K}$ )	Volume ( $V$ , in $\text{L}$ )
1	150	0.30
2	300	0.60
3	450	0.90
4	600	1.20

Based on these results, which of the following statements best describes the relationship between the temperature and volume of the gas sample, and identifies the correct proportionality constant, $k$ , for the equation $V = kT$ ?

Temperature and volume are inversely proportional, with

k = 45\text{ L}\cdot\text{K}

.

Temperature and volume are directly proportional, with

k = 0.002\text{ L/K}

.

Temperature and volume are directly proportional, with

k = 500\text{ K/L}

.

Temperature and volume are inversely proportional, with

k = 0.002\text{ L/K}

.

Cevabı ve açıklamayı göster

Cevap: Temperature and volume are directly proportional, with $k = 0.002\text{ L/K}$ .

Cevap

Temperature and volume are directly proportional, with

k = 0.002\text{ L/K}

.

The correct answer states that temperature and volume are directly proportional with a constant of

0.002\text{ L/K}

. This is correct because as temperature (

T

) increases, volume (

V

) increases at a constant ratio. Solving for the proportionality constant in the equation

V = kT

gives

k = V/T = 0.30\text{ L} / 150\text{ K} = 0.002\text{ L/K}

.

Adım Adım Çözüm

1

Determine the type of relationship between temperature (

T

) and volume (

V

).

Observe that as temperature increases, volume also increases. Specifically, when temperature doubles from

150\text{ K}

to

300\text{ K}

, volume doubles from

0.30\text{ L}

to

0.60\text{ L}

. This indicates a direct relationship.

Recognizing whether a relationship is direct or inverse is necessary to select the correct equation format.

2

Calculate the constant of proportionality,

k

, using the direct proportionality equation

V = kT

.

Solve for

k

:

k = V/T

. Using the data from Trial 1,

k = 0.30\text{ L} / 150\text{ K} = 0.002\text{ L/K}

. Verifying with Trial 2,

k = 0.60\text{ L} / 300\text{ K} = 0.002\text{ L/K}

.

This determines the exact numerical value and units for the constant of proportionality.

Anahtar Kavram

Direct and Inverse Proportionality in Data Analysis

Tahmini Süre:1m 15s

Mathematical Analysis and Data Trends

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