All practice questions

Question 1381Question →

Given the matrix $A = \begin{pmatrix} y & 5 \\ 3 & y+2 \end{pmatrix}$ , if the determinant of $A$ is $9$ and $y > 0$ , calculate the value of $y$ .

Show answer & explanation

Answer: 4

Answer

The value of

y

is 4.

Expanding the determinant of

A

yields

\det(A) = y(y+2) - 15 = y^2 + 2y - 15

. Equating this to 9 gives

y^2 + 2y - 24 = 0

, which factors as

(y+6)(y-4) = 0

. The roots are

y = -6

and

y = 4

. Given that

y > 0

, the required value is 4.

Step-by-Step Solution

1

Find the expression for the determinant of matrix

A

\det(A) = y(y+2) - (5)(3) = y^2 + 2y - 15

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is given by

ad - bc

.

2

Form and solve the quadratic equation

y^2 + 2y - 15 = 9 \implies y^2 + 2y - 24 = 0 \implies (y + 6)(y - 4) = 0

Set the determinant expression equal to the given determinant value of 9 and rearrange into standard quadratic form.

3

Apply the given domain restriction

y > 0

y = 4

The root

y = -6

is discarded because

y

must be strictly positive.

Key Concept

Determinant of a 2x2 matrix and solving non-linear determinant equations

Question 1382Question →

On a topographical map extract drawn at a scale of $1:50,000$ , the total length of all stream channels within a river basin is measured as $18\text{ cm}$ , while the total area of the basin on the map is $12\text{ cm}^2$ . What is the drainage density of this river basin in $\text{km/km}^2$ ?

Show answer & explanation

Answer: 3

Answer

The drainage density of the river basin is

3\text{ km/km}^2

.

Drainage density is determined by dividing total real stream length by total real area of the drainage basin. Given a map scale of

1:50,000

,

18\text{ cm}

of stream length translates to

9\text{ km}

, and

12\text{ cm}^2

of area translates to

3\text{ km}^2

. The resulting quotient gives a drainage density of

3\text{ km/km}^2

.

Step-by-Step Solution

1

Convert linear map measurement to actual distance in kilometers

18\text{ cm} \text{ on map} = 18 \times 0.5\text{ km} = 9\text{ km}

At a scale of

1:50,000

,

1\text{ cm}

represents

50,000\text{ cm}

or

0.5\text{ km}

on the ground.

2

Convert map area measurement to actual area in square kilometers

12\text{ cm}^2 \text{ on map} = 12 \times (0.5\text{ km})^2 = 12 \times 0.25\text{ km}^2 = 3\text{ km}^2

Area scale ratio is the square of the linear scale ratio (

(0.5\text{ km/cm})^2 = 0.25\text{ km}^2/\text{cm}^2

).

3

Calculate drainage density using the formula

D_d = \frac{\sum L}{A}

D_d = \frac{9\text{ km}}{3\text{ km}^2} = 3\text{ km/km}^2

Drainage density is defined as the total length of streams per unit drainage area.

Key Concept

Drainage Density and Scale Conversion

Estimated Time:1m 30s

Question 1383Question →

At a regional agricultural exhibition, $150$ farmers registered their crop cultivation. $85$ farmers grow cassava ( $C$ ), $70$ grow maize ( $M$ ), and $60$ grow yam ( $Y$ ). $35$ farmers grow both cassava and maize, $25$ grow both maize and yam, and $30$ grow both cassava and yam. If $15$ farmers grow all three crops, how many farmers grow none of these three crops?

Show answer & explanation

Answer: 10

Answer

The number of farmers who grow none of the three crops is 10.

Using the Principle of Inclusion-Exclusion for three sets, the number of farmers growing at least one of cassava, maize, or yam is given by

85 + 70 + 60 - 35 - 25 - 30 + 15 = 140

. Since there are

150

farmers in total, the number of farmers who grow none of these three crops is

150 - 140 = 10

.

Step-by-Step Solution

1

Calculate the cardinality of the union of the three sets using the Principle of Inclusion-Exclusion.

|C \cup M \cup Y| = 85 + 70 + 60 - 35 - 25 - 30 + 15 = 140

Summing the three individual set counts overcounts elements in pairwise intersections, which must be subtracted. The central triple intersection is then added back because it was subtracted once too often.

2

Subtract the size of the union from the size of the universal set.

|(C \cup M \cup Y)'| = 150 - 140 = 10

The complement of the union represents the set of farmers outside all three crop categories.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 1384Question →

At a meteorological station in Jos, Nigeria, a weather observer records a maximum daily air temperature of $28.5^\circ\text{C}$ and a minimum daily air temperature of $17.5^\circ\text{C}$ using a Six's maximum and minimum thermometer. What is the mean daily temperature in $^\circ\text{C}$ ?

Show answer & explanation

Answer: 23

Answer

The mean daily temperature is

23.0^\circ\text{C}

.

The mean daily temperature is determined by taking the average of the daily maximum and minimum temperatures:

\frac{28.5^\circ\text{C} + 17.5^\circ\text{C}}{2} = 23.0^\circ\text{C}

.

Step-by-Step Solution

1

Sum the maximum daily temperature and minimum daily temperature recorded at the weather station.

28.5^\circ\text{C} + 17.5^\circ\text{C} = 46.0^\circ\text{C}

Finding the total combined temperature extreme values is the first step in calculating the arithmetic mean.

2

Divide the combined total temperature by 2.

\frac{46.0^\circ\text{C}}{2} = 23.0^\circ\text{C}

The mean daily temperature represents the average temperature over a 24-hour observation period.

Key Concept

Calculation of Mean Daily Temperature

Estimated Time:45s

Question 1385Question →

If $(x, y)$ is a pair of positive integers satisfying the simultaneous equations $x + 2y = 7$ and $x^2 + y^2 = 13$ , what is the value of $2x - y$ ?

Show answer & explanation

Answer: 4

Answer

The value of

2x - y

is 4.

Substituting

x = 7 - 2y

into

x^2 + y^2 = 13

yields the quadratic

5y^2 - 28y + 36 = 0

, which factors as

(5y - 18)(y - 2) = 0

. Since

x

and

y

must both be positive integers, the valid solution pair is

(x, y) = (3, 2)

. Evaluating

2x - y

gives

2(3) - 2 = 4

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation

x + 2y = 7

.

x = 7 - 2y

Isolating

x

allows substitution into the quadratic equation.

2

Substitute

x = 7 - 2y

into

x^2 + y^2 = 13

and expand.

(7 - 2y)^2 + y^2 = 13 \implies 49 - 28y + 4y^2 + y^2 = 13

This reduces the system to a single quadratic equation in terms of

y

.

3

Rearrange and solve the resulting quadratic equation

5y^2 - 28y + 36 = 0

.

(5y - 18)(y - 2) = 0 \implies y = 2

or

y = 3.6

Factoring provides the possible values for

y

.

4

Identify the integer solution pair and compute the target expression

2x - y

.

Since

y

must be a positive integer,

y = 2

and

x = 3

. Therefore,

2x - y = 2(3) - 2 = 4

.

The non-integer solution is rejected based on the positive integer constraint.

Key Concept

Simultaneous Linear and Quadratic Equations

Estimated Time:1m 30s

Question 1386Question →

A candidate takes a two-stage driving examination consisting of a theoretical test followed by a practical test. The probability that the candidate passes the theoretical test is $0.75$ . If the candidate passes the theoretical test, the probability of passing the practical test is $0.80$ . However, if the candidate fails the theoretical test, the probability of passing the practical test is $0.20$ . What is the probability that the candidate passes exactly one of the two tests?

Show answer & explanation

Answer: 0.2

Answer

The probability that the candidate passes exactly one of the two tests is 0.2.

To pass exactly one test, the candidate must either pass theory and fail practical (

0.75 \times 0.20 = 0.15

) or fail theory and pass practical (

0.25 \times 0.20 = 0.05

). Since these two outcomes cannot happen at the same time, their probabilities are added:

0.15 + 0.05 = 0.20

.

Step-by-Step Solution

1

Calculate the probability of passing the theoretical test and failing the practical test.

P(\text{Theory} \cap \text{Practical}') = 0.75 \times (1 - 0.80) = 0.15

The probability of failing practical given passing theory is

1 - 0.80 = 0.20

.

2

Calculate the probability of failing the theoretical test and passing the practical test.

P(\text{Theory}' \cap \text{Practical}) = (1 - 0.75) \times 0.20 = 0.05

The probability of failing theory is

1 - 0.75 = 0.25

.

3

Sum the probabilities of the two mutually exclusive outcomes.

0.15 + 0.05 = 0.20

Passing exactly one test is the union of two disjoint compound events.

Key Concept

Compound probability laws and conditional independence structure in sequential events

Estimated Time:1m 30s

Question 1387Question →

A constant-volume gas thermometer registers a pressure of $60\,\text{kPa}$ at the ice point ( $0^\circ\text{C}$ ) and $84\,\text{kPa}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the pressure registered by the thermometer is $72\,\text{kPa}$ ?

Show answer & explanation

Answer: 50

Answer

The temperature corresponding to a pressure reading of

72\,\text{kPa}

is

50^\circ\text{C}

.

Using the linear relation for a constant-volume gas thermometer:

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

. Substituting

P_T = 72\,\text{kPa}

,

P_0 = 60\,\text{kPa}

, and

P_{100} = 84\,\text{kPa}

yields

T = \frac{72 - 60}{84 - 60} \times 100 = \frac{12}{24} \times 100 = 50^\circ\text{C}

.

Step-by-Step Solution

1

Identify the thermometric property values at the fixed points and target state.

P_0 = 60\,\text{kPa}

,

P_{100} = 84\,\text{kPa}

, and

P_T = 72\,\text{kPa}

.

These represent the pressure values corresponding to

0^\circ\text{C}

,

100^\circ\text{C}

, and the unknown temperature

T

respectively.

2

Set up the linear scale conversion equation.

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

Temperature changes linearly with the thermometric property (gas pressure at constant volume).

3

Calculate the numerical value.

T = \frac{12}{24} \times 100 = 50^\circ\text{C}

Simplifying the fraction

\frac{12}{24} = 0.5

and multiplying by

100

gives

50

.

Key Concept

Temperature measurement using constant-volume gas pressure as a thermometric property

Question 1388Question →

In $\triangle ABC$ , side $b = 10\text{ cm}$ , side $c = 6\text{ cm}$ , and $\angle A = 120^\circ$ . What is the length of side $a$ in centimeters?

Show answer & explanation

Answer: 14

Answer

The length of side

a

is

14\text{ cm}

.

Applying the Cosine Rule

a^2 = b^2 + c^2 - 2bc \cos A

with

b = 10\text{ cm}

,

c = 6\text{ cm}

, and

\angle A = 120^\circ

gives

a^2 = 10^2 + 6^2 - 2(10)(6)(-0.5) = 100 + 36 + 60 = 196

. Taking the square root yields

a = 14\text{ cm}

.

Step-by-Step Solution

1

Identify the given parameters and select the relevant trigonometric rule

Given sides

b = 10\text{ cm}

,

c = 6\text{ cm}

, and included angle

\angle A = 120^\circ

. Use the Cosine Rule:

a^2 = b^2 + c^2 - 2bc \cos A

.

The Cosine Rule is required because two sides and the included angle (SAS) are given to find the third side.

2

Substitute the known values into the Cosine Rule equation

a^2 = 10^2 + 6^2 - 2(10)(6) \cos(120^\circ)

This sets up a single equation with the unknown side length

a

.

3

Evaluate the cosine term and simplify the arithmetic expression

a^2 = 100 + 36 - 120(-0.5) = 136 + 60 = 196

Since

120^\circ

is an obtuse angle in the second quadrant,

\cos(120^\circ) = -0.5

, changing the minus sign in the formula to a plus sign.

4

Calculate the principal square root to find

a

a = \sqrt{196} = 14\text{ cm}

Side length must be a positive scalar quantity.

Key Concept

Cosine Rule for Side Length in Oblique Triangles

Question 1389Question →

In a nuclear fusion process, two deuterium nuclei (

\text{^{2}_{1}H}

) fuse to form a helium-3 nucleus (

\text{^{3}_{2}He}

) and a neutron (

\text{^{1}_{0}n}

) according to the reaction equation:

\text{^{2}_{1}H} + \text{^{2}_{1}H} \rightarrow \text{^{3}_{2}He} + \text{^{1}_{0}n} + Q

Given the mass values:
- Mass of $\text{^{2}_{1}H} = 2.0141\text{ u}$
- Mass of $\text{^{3}_{2}He} = 3.0160\text{ u}$
- Mass of $\text{^{1}_{0}n} = 1.0087\text{ u}$

Using the conversion factor $1\text{ u} = 931.5\text{ MeV}$ , what is the energy released ( $Q$ ) in this fusion reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 3.26

Answer

The total energy released (

Q

) in the reaction is

3.26\text{ MeV}

.

The energy released in a nuclear fusion reaction is proportional to the decrease in total rest mass (mass defect). Summing the mass of two deuterium nuclei gives

4.0282\text{ u}

, while the sum of the product masses (

\text{^{3}_{2}He}

and a neutron) is

4.0247\text{ u}

. Subtracting these yields a mass defect of

0.0035\text{ u}

. Multiplying

0.0035\text{ u}

by

931.5\text{ MeV/u}

gives

3.26\text{ MeV}

of released energy.

Step-by-Step Solution

1

Calculate the total initial mass of the reacting deuterium nuclei.

m_{\text{reactants}} = 2 \times 2.0141\text{ u} = 4.0282\text{ u}

Two deuterium nuclei participate on the reactant side of the equation.

2

Calculate the total final mass of the products.

m_{\text{products}} = 3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

The reaction produces one helium-3 nucleus and one neutron.

3

Determine the mass defect (difference between reactant and product masses).

\Delta m = 4.0282\text{ u} - 4.0247\text{ u} = 0.0035\text{ u}

The mass lost during fusion is converted into nuclear kinetic energy and radiation.

4

Convert the mass defect into energy in MeV using the conversion factor.

Q = 0.0035\text{ u} \times 931.5\text{ MeV/u} = 3.26025\text{ MeV} \approx 3.26\text{ MeV}

Each atomic mass unit lost corresponds to

931.5\text{ MeV}

of energy.

Key Concept

Mass defect and energy release in nuclear fusion reactions (

E = \Delta m c^2

)

Question 1390Question →

In a nuclear fusion reaction, two deuterium nuclei ( ${^{2}_{1}\text{H}}$ ) combine to form a helium-3 nucleus ( ${^{3}_{2}\text{He}}$ ) and a neutron ( ${^{1}_{0}\text{n}}$ ). The total mass of the two reactant deuterium nuclei is $4.0282\text{ u}$ , while the total mass of the resulting helium-3 and neutron products is $4.0247\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , calculate the total energy released in this reaction in $\text{MeV}$ .

Show answer & explanation

Answer: 3.26

Answer

The total energy released in the nuclear fusion reaction is approximately 3.26 MeV.

The energy released in a nuclear fusion reaction is determined by the mass defect, which is the difference between the total mass of reactants and the total mass of products. Subtracting

4.0247\text{ u}

from

4.0282\text{ u}

yields a mass defect of

0.0035\text{ u}

. Multiplying this mass defect by the mass-energy conversion factor of

931.5\text{ MeV/u}

gives an energy release of approximately

3.26\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (Δm)

Δm = 4.0282 u - 4.0247 u = 0.0035 u

Mass defect is the difference between the initial mass of reactants and the final mass of products in a nuclear reaction.

2

Calculate the energy released (E) in MeV

E = 0.0035 u × 931.5 MeV/u = 3.26025 MeV

According to mass-energy equivalence, 1 unified atomic mass unit (u) liberates 931.5 MeV of energy.

Key Concept

Mass defect and energy release in nuclear fusion

Question 1391Question →

The length and width of a rectangular metal sheet are measured as $(8.0 \pm 0.2)\text{ cm}$ and $(5.0 \pm 0.1)\text{ cm}$ , respectively. What is the percentage error in the calculated area of the metal sheet?

Show answer & explanation

Answer: 4.5

Answer

The percentage error in the calculated area of the metal sheet is

4.5\%

.

For a calculated quantity involving multiplication (

A = L \times W

), the total relative error is the sum of the relative errors of the individual measurements. The fractional error in length is

\frac{0.2}{8.0} = 0.025

(

2.5\%

) and in width is

\frac{0.1}{5.0} = 0.020

(

2.0\%

). Summing these gives

0.045

, which is equivalent to

4.5\%

.

Step-by-Step Solution

1

Determine the fractional error in the length measurement

\frac{0.2\text{ cm}}{8.0\text{ cm}} = 0.025

Fractional error is given by the ratio of absolute error to the measured value.

2

Determine the fractional error in the width measurement

\frac{0.1\text{ cm}}{5.0\text{ cm}} = 0.020

Fractional error is calculated as the absolute uncertainty divided by the measured dimension.

3

Combine the fractional errors for the calculated area

\frac{\Delta A}{A} = 0.025 + 0.020 = 0.045

For quantities multiplied together (

A = L \times W

), individual fractional errors sum to give the total fractional error.

4

Convert the fractional error to percentage error

0.045 \times 100\% = 4.5\%

Multiplying the relative error by 100 yields the percentage error.

Key Concept

Error Propagation in Products

Question 1392Question →

A swimming pool has an apparent depth of $1.8\text{ m}$ when viewed vertically from directly above. If the refractive index of water relative to air is $\frac{4}{3}$ , what is the real depth of the pool in meters?

Show answer & explanation

Answer: 2.4

Answer

The real depth of the pool is

2.4\text{ m}

.

The refractive index

n

of a medium is defined as the ratio of the real depth to the apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Multiplying the observed apparent depth of

1.8\text{ m}

by the refractive index

\frac{4}{3}

gives the true real depth of

2.4\text{ m}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

Refractive index

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

.

Light bending at the boundary causes an object submerged in a denser medium to appear closer to the surface.

2

Rearrange the equation to make Real Depth the subject.

\text{Real Depth} = n \times \text{Apparent Depth}.

To calculate the true depth from the observed apparent depth and the optical density of water.

3

Substitute the given values into the formula.

\text{Real Depth} = \frac{4}{3} \times 1.8\text{ m} = 2.4\text{ m}.

Multiplying the apparent depth by the refractive index yields the actual physical depth.

Key Concept

Refraction of Light and Real/Apparent Depth

Question 1393Question →

A sample of a pure iron oxide synthesized in a laboratory contains $5.60\text{ g}$ of iron and $2.40\text{ g}$ of oxygen. According to the Law of Definite Proportions, what is the mass of iron, in grams, present in a $20.0\text{ g}$ sample of the same iron oxide collected from a natural deposit?

Show answer & explanation

Answer: 14

Answer

The mass of iron present in the 20.0 g sample of iron oxide is 14.0 g.

According to the Law of Definite Proportions, a pure chemical compound always contains the same elements combined together in the exact same proportion by mass, regardless of its source or method of preparation. In the laboratory sample,

8.00\text{ g}

of iron oxide contains

5.60\text{ g}

of iron, giving an iron mass composition of

70\%

. Therefore, a

20.0\text{ g}

natural sample of the same compound must also contain

70\%

iron by mass, which equals

14.0\text{ g}

.

Step-by-Step Solution

1

Calculate total mass of the laboratory sample

5.60\text{ g} + 2.40\text{ g} = 8.00\text{ g}

The total mass of the compound is the sum of the constituent element masses (Law of Conservation of Mass).

2

Find the mass percentage/fraction of iron

\frac{5.60\text{ g}}{8.00\text{ g}} = 0.70

(or

70\%

)

By the Law of Definite Proportions, the mass ratio of elements in a pure compound is constant.

3

Calculate mass of iron in the 20.0 g natural sample

0.70 \times 20.0\text{ g} = 14.0\text{ g}

Applying the constant mass composition percentage to the new sample mass.

Key Concept

Law of Definite Proportions (Constant Composition)

Estimated Time:1m 30s

Question 1394Question →

A stopwatch with a zero error of $-0.25\text{ s}$ is used to record the time taken for an object to travel down an inclined path. If the stopwatch displays a time reading of $14.65\text{ s}$ at the end of the motion, what is the true time elapsed in seconds?

Show answer & explanation

Answer: 14.9

Answer

The true time elapsed is 14.90 seconds.

The correct answer is obtained by subtracting the zero error from the observed reading. Since the stopwatch has a negative zero error of

-0.25\text{ s}

, the true time is

14.65\text{ s} - (-0.25\text{ s}) = 14.90\text{ s}

.

Step-by-Step Solution

1

Apply the zero error correction formula for measurement instruments

Actual Time = Displayed Time - Zero Error

Zero error represents a constant bias on the instrument that must be subtracted from the uncorrected reading.

2

Substitute the given values into the formula

Actual Time = 14.65 s - (-0.25 s)

The instrument starts behind zero by 0.25 s, so the zero error value is negative.

3

Perform the calculation

Actual Time = 14.90 s

Subtracting a negative quantity is mathematically equivalent to adding its positive magnitude.

Key Concept

Zero Error Correction in Stopwatch Time Measurement

Question 1395Question →

An object undergoing simple harmonic motion completes $20$ complete oscillations in a time duration of $10.0\text{ s}$ . What is the period of oscillation of the object in seconds?

Show answer & explanation

Answer: 0.5

Answer

The period of oscillation is

0.5\text{ s}

.

The period of simple harmonic motion is the time taken to complete one single oscillation. Dividing the total time (

10.0\text{ s}

) by the number of oscillations (

20

) yields

0.5\text{ s}

.

Step-by-Step Solution

1

Apply the definition of oscillation period

T = \frac{t}{N}

Period

T

measures the time required for a single complete cycle.

2

Calculate the numerical value for period

T = \frac{10.0\text{ s}}{20} = 0.5\text{ s}

Dividing total elapsed time by total completed oscillations gives time per oscillation.

Key Concept

Period of Simple Harmonic Motion

Question 1396Question →

How much energy, in $\text{MeV}$ , is released when a nuclear fission process results in a mass defect of $0.05 \text{ u}$ ? (Take $1 \text{ u} = 931.5 \text{ MeV}$ )

Show answer & explanation

Answer: 46.575

Answer

46.575 MeV

The total energy released in a nuclear fission reaction is found by multiplying the mass defect by the energy equivalent of 1 atomic mass unit. Multiplying

0.05 \text{ u}

by

931.5 \text{ MeV/u}

gives

46.575 \text{ MeV}

.

Step-by-Step Solution

1

Identify the mass defect and the mass-to-energy conversion factor.

Mass defect

\Delta m = 0.05 \text{ u}

, and

1 \text{ u} = 931.5 \text{ MeV}

.

In nuclear fission, mass lost during the reaction is converted directly into energy according to Einstein's mass-energy equivalence principle.

2

Calculate the total energy released.

E = 0.05 \text{ u} \times 931.5 \text{ MeV/u} = 46.575 \text{ MeV}

.

Multiplying the mass defect in atomic mass units by

931.5 \text{ MeV/u}

gives the total released energy in

\text{MeV}

.

Key Concept

Mass-Energy Conversion in Nuclear Reactions

Question 1397Question →

A conveyor system pulls a $40\text{ kg}$ crate at a constant speed of $3\text{ m s}^{-1}$ up a rough inclined ramp. The ramp rises $3\text{ m}$ for every $5\text{ m}$ measured along its slope (giving $\sin\theta = 0.6$ and $\cos\theta = 0.8$ ). If the coefficient of kinetic friction between the crate and the ramp is $0.25$ and $g = 10\text{ m s}^{-2}$ , what is the power output of the conveyor system in watts?

Show answer & explanation

Answer: 960

Answer

The power output required by the conveyor system is

960\text{ W}

.

To pull the crate up the ramp at constant speed, the conveyor force must overcome both the parallel gravitational component (

240\text{ N}

) and friction (

80\text{ N}

), making the total force

320\text{ N}

. Multiplying this force by the constant speed of

3\text{ m s}^{-1}

gives a total power output of

960\text{ W}

.

Step-by-Step Solution

1

Calculate the component of weight parallel to the inclined plane

F_g = mg \sin\theta = 40 \times 10 \times 0.6 = 240\text{ N}

Gravity pulls the object back down along the slope with force

mg \sin\theta

.

2

Calculate the normal reaction force perpendicular to the plane

N = mg \cos\theta = 40 \times 10 \times 0.8 = 320\text{ N}

The normal force balances the perpendicular weight component.

3

Determine the magnitude of kinetic friction force

f_k = \mu N = 0.25 \times 320 = 80\text{ N}

Friction opposes motion up the slope and depends on the normal force.

4

Calculate the total pulling force needed for zero net acceleration

F = F_g + f_k = 240 + 80 = 320\text{ N}

At constant velocity, net force along the incline must equal zero, so

F = mg\sin\theta + f_k

.

5

Calculate the power output of the conveyor

P = F \times v = 320\text{ N} \times 3\text{ m s}^{-1} = 960\text{ W}

Power developed by a constant force moving an object at velocity

v

is given by

P = Fv

.

Key Concept

Work done against gravity and friction, and rate of doing work (Power

P = Fv

)

Question 1398Question →

Consider the following standard enthalpies of combustion at

298\text{ K}

:

\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -394\text{ kJ mol}^{-1}

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -286\text{ kJ mol}^{-1}

\text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(l) \quad \Delta H^\circ = -1300\text{ kJ mol}^{-1}

Using Hess's law, calculate the standard enthalpy of formation of ethyne gas, $\text{C}_2\text{H}_2(g)$ , in $\text{kJ mol}^{-1}$ .

Show answer & explanation

Answer: 226

Answer

The standard enthalpy of formation of ethyne gas is

+226\text{ kJ mol}^{-1}

(or

226\text{ kJ mol}^{-1}

).

Applying Hess's law involves expressing the enthalpy of formation of ethyne as the sum of twice the enthalpy of combustion of carbon, once the enthalpy of combustion of hydrogen, minus the enthalpy of combustion of ethyne:

\Delta H_f^\circ = 2(-394) + (-286) - (-1300) = -788 - 286 + 1300 = +226\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Write the standard formation equation for ethyne

2\text{C}(s) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_2(g)

The enthalpy of formation represents the formation of one mole of a compound from its constituent elements in their standard states.

2

Manipulate given thermochemical equations to match the target equation

Multiply equation 1 by 2 (

\Delta H = -788\text{ kJ}

), keep equation 2 unchanged (

\Delta H = -286\text{ kJ}

), and reverse equation 3 (

\Delta H = +1300\text{ kJ}

)

According to Hess's Law, changing stoichiometric coefficients multiplies

\Delta H

by the same factor, and reversing a reaction flips the sign of

\Delta H

.

3

Sum the enthalpy values for the target reaction

\Delta H_f^\circ = -788 - 286 + 1300 = 226\text{ kJ mol}^{-1}

The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for each intermediate step.

Key Concept

Hess's Law of Constant Heat Summation

Question 1399Question →

A wheel and axle system consists of a wheel with a radius of $25\text{ cm}$ attached to an axle with a radius of $5\text{ cm}$ . What is the velocity ratio of this machine?

Show answer & explanation

Answer: 5

Answer

The velocity ratio of the machine is 5.

The velocity ratio (VR) of a wheel and axle is the ratio of the radius of the wheel (

R

) to the radius of the axle (

r

). Calculating

\frac{25\text{ cm}}{5\text{ cm}}

gives a velocity ratio of

5

.

Step-by-Step Solution

1

Identify the formula for the velocity ratio of a wheel and axle system.

\text{VR} = \frac{R}{r}

Velocity ratio is defined as the distance moved by the effort (proportional to wheel radius) divided by the distance moved by the load (proportional to axle radius).

2

Substitute the given values into the formula.

\text{VR} = \frac{25\text{ cm}}{5\text{ cm}}

The radius of the wheel

R = 25\text{ cm}

and the radius of the axle

r = 5\text{ cm}

.

3

Calculate the final ratio.

\text{VR} = 5

Dividing 25 by 5 yields 5. The ratio is dimensionless because the units of centimeters cancel out.

Key Concept

Velocity Ratio of a Wheel and Axle

Question 1400Question →

A uniform horizontal wooden rod $XY$ of length $3.0\text{ m}$ and weight $50\text{ N}$ rests horizontally on two smooth supports located at $X$ (the left end) and at a point $Z$ which is $0.6\text{ m}$ from end $Y$ . A block of weight $120\text{ N}$ is placed on the rod at a distance of $0.9\text{ m}$ from end $X$ . What is the magnitude of the upward reaction force, in newtons, at support $Z$ ?

Show answer & explanation

Answer: 76.25

Answer

The magnitude of the upward reaction force at support

Z

is

76.25\text{ N}

.

Taking moments about support

X

, the total clockwise moment is the sum of the moment due to the load (

120\text{ N} \times 0.9\text{ m} = 108\text{ N}\cdot\text{m}

) and the weight of the rod (

50\text{ N} \times 1.5\text{ m} = 75\text{ N}\cdot\text{m}

), giving

183\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the reaction force at

Z

(

R_Z \times 2.4\text{ m}

) yields

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

Step-by-Step Solution

1

Identify the perpendicular distance of each force and support from pivot point

X

.

Center of gravity position

x_{cg} = 1.5\text{ m}

, load position

x_{L} = 0.9\text{ m}

, and support

Z

position

x_{Z} = 3.0 - 0.6 = 2.4\text{ m}

.

Taking moments about

X

requires knowing the exact moment arm for each force from

X

.

2

Set up the equation for rotational equilibrium about point

X

.

Total clockwise moment =

(120 \times 0.9) + (50 \times 1.5) = 183\text{ N}\cdot\text{m}

; Total counterclockwise moment =

R_Z \times 2.4

.

Choosing pivot

X

eliminates the unknown reaction force

R_X

because its distance from

X

is zero.

3

Equate clockwise moments to counterclockwise moments and solve for

R_Z

.

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

For a body in rotational equilibrium, the algebraic sum of moments about any point must equal zero.

Key Concept

Principle of Moments and Rotational Equilibrium

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