All practice questions

Question 7721Question →

Given the matrix $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}$ , if $\det(A) = 10$ , what is the value of $k$ ?

Show answer & explanation

Answer: -3

Answer

The value of

k

is

-3

.

Expanding the determinant along the first row gives

1(4k) - 2(3k - 2) + 0 = -2k + 4

. Setting

-2k + 4 = 10

leads directly to

-2k = 6

, giving

k = -3

.

Step-by-Step Solution

1

Expand the determinant of matrix

A

along the first row.

\det(A) = 1(4k - 0) - 2(3k - 2) + 0 = -2k + 4

Using cofactor expansion along the top row to find the expression for the determinant.

2

Set the calculated determinant equal to the given value and solve for

k

.

-2k + 4 = 10 \implies -2k = 6 \implies k = -3

Equating the determinant algebraic expression to 10.

Key Concept

Determinant of a 3x3 Matrix

Question 7722Question →

When the polynomial $P(x) = x^3 + ax^2 + bx - 6$ is divided by $(x - 2)$ , the remainder is $0$ . When $P(x)$ is divided by $(x + 1)$ , the remainder is $12$ . What is the value of $a + b$ ?

-7

17

7

19

Show answer & explanation

Answer: $-7$

Answer

The value of

a + b

is

-7

.

According to the Remainder Theorem, dividing

P(x)

by

(x - 2)

with a remainder of

0

means

P(2) = 0

, giving the equation

2a + b = -1

. Dividing

P(x)

by

(x + 1)

with a remainder of

12

means

P(-1) = 12

, giving

a - b = 19

. Solving these two linear equations simultaneously yields

a = 6

and

b = -13

. Adding these values together gives

a + b = -7

.

Step-by-Step Solution

1

Apply the Factor/Remainder Theorem for divisor

(x - 2)

P(2) = 2^3 + a(2)^2 + b(2) - 6 = 0 \implies 4a + 2b + 2 = 0 \implies 2a + b = -1

Since dividing

P(x)

by

(x - 2)

leaves a remainder of

0

,

P(2) = 0

.

2

Apply the Remainder Theorem for divisor

(x + 1)

P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 6 = 12 \implies a - b - 7 = 12 \implies a - b = 19

Setting the linear divisor

x + 1 = 0

gives

x = -1

, so

P(-1) = 12

.

3

Solve the system of simultaneous linear equations for

a

and

b

Adding

(2a + b = -1)

and

(a - b = 19)

yields

3a = 18 \implies a = 6

. Substituting

a = 6

into

a - b = 19

gives

6 - b = 19 \implies b = -13

.

Eliminating

b

allows finding the values of constants

a

and

b

.

4

Calculate

a + b

a + b = 6 + (-13) = -7

Summing the determined constants

a

and

b

gives the target expression.

Key Concept

Remainder and Factor Theorems

Estimated Time:1m 30s

Question 7723Question →

The area of the region bounded by the curve $y = 3x^2 - 4x + 3$ , the $x$ -axis, and the vertical lines $x = 0$ and $x = k$ (where $k > 0$ ) is $18$ square units. What is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of

k

is

3

.

The area under the non-negative curve

y = 3x^2 - 4x + 3

from

x = 0

to

x = k

is found by calculating

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k

. Setting this expression equal to

18

gives

k^3 - 2k^2 + 3k - 18 = 0

. By the factor theorem, testing

k = 3

yields

3^3 - 2(3)^2 + 3(3) - 18 = 0

. Factoring as

(k - 3)(k^2 + k + 6) = 0

confirms

k = 3

as the only real solution.

Step-by-Step Solution

1

Set up the definite integral for the area bounded by the curve and the x-axis

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18

The curve

y = 3x^2 - 4x + 3

lies entirely above the x-axis for all real

x

because its leading coefficient is positive and its discriminant

(-4)^2 - 4(3)(3) = -20 < 0

.

2

Evaluate the definite integral in terms of

k

\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k

Applying the fundamental theorem of calculus by integrating term by term.

3

Form and simplify the polynomial equation

k^3 - 2k^2 + 3k - 18 = 0

Equating the definite integral expression to the given area value of 18.

4

Solve for real values of

k > 0

k = 3

Using the factor theorem on

k^3 - 2k^2 + 3k - 18 = 0

,

k = 3

yields zero (

27 - 18 + 9 - 18 = 0

). Factoring out

(k - 3)

gives

(k - 3)(k^2 + k + 6) = 0

, where

k^2 + k + 6 = 0

has complex roots.

Key Concept

Determining an unknown boundary limit of a definite integral representing area under a curve

Question 7724Question →

Calculate the gradient of the normal line to the curve $y = \frac{6}{x}$ at the point where $x = 3$ .

Show answer & explanation

Answer: 1.5

Answer

The gradient of the normal line is 1.5.

Differentiating

y = 6x^{-1}

yields

\frac{dy}{dx} = -\frac{6}{x^2}

. Evaluating this derivative at

x = 3

gives the tangent gradient

m_t = -\frac{6}{9} = -\frac{2}{3}

. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal

m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5

.

Step-by-Step Solution

1

Differentiate the function

y = 6x^{-1}

with respect to

x

\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}

The first derivative represents the formula for the tangent gradient to the curve at any given point.

2

Evaluate the derivative at

x = 3

to find the tangent slope (

m_t

)

m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}

Substituting the given x-coordinate into the derivative gives the exact slope of the tangent at that point.

3

Calculate the normal slope (

m_n

) as the negative reciprocal of

m_t

m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5

The normal line is perpendicular to the tangent line, meaning

m_t \cdot m_n = -1

.

Key Concept

The gradient of the normal to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative evaluated at that point:

m_n = -\frac{1}{f'(a)}

.

Question 7725Question →

A pie chart illustrates the distribution of undergraduate students enrolled across four faculties at a university: Arts, Science, Law, and Medicine. The central angles for the sectors representing Arts, Science, and Law are $120^\circ$ , $90^\circ$ , and $75^\circ$ , respectively. If $300$ students are enrolled in the Faculty of Medicine, what is the total number of students enrolled in the university?

Show answer & explanation

Answer: 1440

Answer

The total number of students enrolled in the university is 1440.

The total sum of central angles in any pie chart is

360^\circ

. Subtracting the given angles for Arts (

120^\circ

), Science (

90^\circ

), and Law (

75^\circ

) from

360^\circ

gives the sector angle for Medicine as

75^\circ

. Since

75^\circ

represents

300

students, each degree represents

\frac{300}{75} = 4

students. Multiplying

4

students per degree by the total

360^\circ

yields

1440

total students in the university.

Step-by-Step Solution

1

Determine the sector angle for Medicine.

Sector angle for Medicine =

75^\circ

The sum of central angles in a pie chart is

360^\circ

. Subtracting

120^\circ + 90^\circ + 75^\circ = 285^\circ

from

360^\circ

gives

75^\circ

.

2

Formulate the proportion relating sector angle to frequency.

\frac{75^\circ}{360^\circ} \times N = 300

, where

N

represents total students.

The fractional portion of the angle (

75^\circ

out of

360^\circ

) equals the fractional portion of the total student count (

300

out of

N

).

3

Calculate the total student population

N

.

N = \frac{300 \times 360}{75} = 1440

Dividing

300

by

75

yields

4

students per degree. Multiplying

4

by

360

gives

1440

students.

Key Concept

Pie Chart Sector Angle and Total Population Calculation

Question 7726Question →

The tensile stress $\sigma$ on a solid wire subjected to a stretching force $F$ is defined as force per unit cross-sectional area, while the fractional change in length is the tensile strain $\epsilon$ . If Young's modulus $Y$ of the material is given by $Y = \frac{\sigma}{\epsilon}$ , which of the following is the SI unit of Young's modulus expressed in fundamental SI base units?

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^{-2}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}

Show answer & explanation

Answer: $\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Answer

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Young's modulus is calculated as tensile stress divided by tensile strain. Since tensile strain is the ratio of change in length to original length, it has no units. Therefore, the SI unit of Young's modulus is identical to that of stress. Stress is force divided by area:

\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Step-by-Step Solution

1

Determine the dimensions of the force component

Force

F = ma \implies [F] = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Mass has fundamental unit

\text{kg}

and acceleration has derived unit

\text{m}\cdot\text{s}^{-2}

.

2

Determine the SI base units of tensile stress

\sigma

[\sigma] = \frac{[F]}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Stress is defined as force divided by cross-sectional area (

A = \text{m}^2

).

3

Evaluate the unit of Young's modulus

Y

[Y] = \frac{[\sigma]}{[\epsilon]} = \frac{\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}}{1} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Strain

\epsilon = \frac{\Delta L}{L}

is the ratio of two lengths and is dimensionless.

Key Concept

Derivation of SI base units for mechanical moduli from physical definitions

Estimated Time:1m 15s

Question 7727Question →

In a mercury vapor tube, an excited atom transitions from an upper energy level of $-3.71\text{ eV}$ to a lower energy level of $-5.54\text{ eV}$ . What is the frequency of the emitted photon? (Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

4.44 \times 10^{14}\text{ Hz}

2.24 \times 10^{15}\text{ Hz}

1.34 \times 10^{15}\text{ Hz}

2.77 \times 10^{33}\text{ Hz}

Show answer & explanation

Answer: $4.44 \times 10^{14}\text{ Hz}$

Answer

The frequency of the emitted photon is

4.44 \times 10^{14}\text{ Hz}

.

The energy lost by the atom during transition is

\Delta E = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}

. Converting to Joules gives

1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant (

6.6 \times 10^{-34}\text{ J s}

) yields a photon frequency of

4.44 \times 10^{14}\text{ Hz}

.

Step-by-Step Solution

1

Calculate the energy difference (

\Delta E

) between the two atomic energy levels.

\Delta E = E_{\text{initial}} - E_{\text{final}} = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}

The energy of the emitted photon equals the difference in energy between the initial and final states.

2

Convert the energy difference from electron-volts (eV) to Joules (J).

\Delta E = 1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}

Planck's constant is given in SI units (J s), so the energy must be in Joules.

3

Apply Planck's equation

E = hf

to find the photon frequency

f

.

f = \frac{\Delta E}{h} = \frac{2.928 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}} \approx 4.44 \times 10^{14}\text{ Hz}

The frequency of an emitted photon is directly proportional to its energy difference.

Key Concept

Atomic transition photon frequency calculation (

E = hf = E_2 - E_1

).

Estimated Time:1m 30s

Question 7728Question →

Under which of the following conditions will total internal reflection occur when light encounters the boundary between two transparent media?

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence less than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at normal incidence.

Show answer & explanation

Answer: When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

Answer

Total internal reflection occurs when light travels from a medium of higher refractive index to a medium of lower refractive index at an angle of incidence greater than the critical angle.

Total internal reflection takes place only when light moves from an optically denser medium (higher refractive index) into an optically less dense medium (lower refractive index), and the angle of incidence at the interface is greater than the critical angle for the two media.

Step-by-Step Solution

1

Identify the optical density requirement for total internal reflection

Light must travel from an optically denser medium (higher refractive index

n_1

) toward an optically rarer medium (lower refractive index

n_2

).

This condition ensures that the light refracts away from the normal into the second medium.

2

Identify the angular requirement at the interface

The angle of incidence

i

must be strictly greater than the critical angle

\theta_c

(

\sin\theta_c = \frac{n_2}{n_1}

).

When

i > \theta_c

, no light can refract into the second medium, resulting in total reflection back into the initial medium.

Key Concept

Conditions for Total Internal Reflection

Question 7729Question →

Given the matrices $A = \begin{pmatrix} x & 3 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ , what is the value of $x$ such that the determinant of the product matrix $AB$ is $25$ ?

4

2

-1

11.5

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

The determinant of a product of matrices equals the product of their determinants,

\det(AB) = \det(A)\det(B)

. Since

\det(B) = 5

and

\det(A) = 2x - 3

, setting

5(2x - 3) = 25

yields

2x - 3 = 5

, giving

x = 4

.

Step-by-Step Solution

1

Calculate the determinant of matrix B

\det(B) = (2)(4) - (1)(3) = 8 - 3 = 5

The determinant of a 2x2 matrix

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

is given by

ad - bc

.

2

Express the determinant of matrix A in terms of x

\det(A) = (x)(2) - (3)(1) = 2x - 3

Apply the 2x2 determinant formula to matrix A.

3

Use the determinant product rule det(AB) = det(A) * det(B)

\det(AB) = (2x - 3) \cdot 5 = 10x - 15

The determinant of the product of two square matrices equals the product of their individual determinants.

4

Set det(AB) equal to 25 and solve for x

10x - 15 = 25 \implies 10x = 40 \implies x = 4

Equate the determinant expression to the given value 25 to isolate x.

Key Concept

Determinant of a Matrix Product

Question 7730Question →

According to Bohr's atomic model of the hydrogen atom, what condition must be satisfied by an electron moving in a stable stationary orbit?

Its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Its total energy is proportional to light intensity.

Its orbital frequency increases continuously as it radiates energy in the orbit.

Its total mechanical energy is zero in all allowed stationary orbits.

Show answer & explanation

Answer: Its orbital angular momentum is an integral multiple of $\frac{h}{2\pi}$ .

Answer

An electron moves in a stable stationary orbit when its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Bohr's fundamental postulate states that an electron can revolve around the nucleus only in certain non-radiating orbits (stationary states) where its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Step-by-Step Solution

1

Recall Bohr's postulates for the hydrogen atom

Identify that stable orbits require quantization of orbital angular momentum.

Bohr introduced quantization to explain why orbiting electrons do not continuously emit radiation and collapse into the nucleus.

2

State the mathematical formula for angular momentum quantization

L = mvr = \frac{nh}{2\pi}

, where

n

is an integer (

1, 2, 3, \dots

) and

h

is Planck's constant.

Only specific discreet orbits meeting this condition are allowed stationary states.

Key Concept

Quantization of Angular Momentum in Bohr's Model

Estimated Time:45s

Question 7731Question →

Given two matrices $A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}$ , where $x > 0$ . If the determinant of the product matrix $AB$ is $20$ , find the value of $x$ .

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Using the property

\det(AB) = \det(A) \cdot \det(B)

, we compute

\det(A) = 3x - 4

and

\det(B) = x + 1

. Equating their product to 20 gives

(3x - 4)(x + 1) = 20

, which simplifies to the quadratic

3x^2 - x - 24 = 0

. Factorizing yields

(3x + 8)(x - 3) = 0

. Since

x

must be a positive number, the correct value is

x = 3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

\det(A) = 3x - 4

The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.

2

Calculate the determinant of matrix B

\det(B) = x + 1

Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.

3

Use the determinant product property det(AB) = det(A) * det(B)

(3x - 4)(x + 1) = 20

The determinant of the product of two square matrices equals the product of their individual determinants.

4

Form and solve the quadratic equation

3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0

Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.

5

Determine the positive solution for x

x = 3

Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Key Concept

Determinant of Matrix Product and 2x2 Determinants

Estimated Time:2m 0s

Question 7732Question →

A frustum of a right circular cone has an upper base radius of $3\text{ cm}$ , a lower base radius of $6\text{ cm}$ , and a vertical height of $4\text{ cm}$ . What is the volume of the frustum?

84\pi\text{ cm}^3

252\pi\text{ cm}^3

60\pi\text{ cm}^3

81\pi\text{ cm}^3

Show answer & explanation

Answer: $84\pi\text{ cm}^3$

Answer

The volume of the frustum is

84\pi\text{ cm}^3

.

The correct answer is obtained by applying the frustum volume formula

V = \frac{1}{3}\pi h (R^2 + r^2 + R r)

. Substituting

R = 6

,

r = 3

, and

h = 4

gives

V = \frac{1}{3}\pi(4)(36 + 9 + 18) = 84\pi\text{ cm}^3

.

Step-by-Step Solution

1

Identify the given dimensions and formula for the volume of a frustum of a right circular cone.

Upper radius

r = 3\text{ cm}

, lower radius

R = 6\text{ cm}

, height

h = 4\text{ cm}

. Formula:

V = \frac{1}{3}\pi h (R^2 + r^2 + R r)

.

The volume of a frustum of a cone is derived by subtracting the top small cone from the total original cone.

2

Evaluate the terms inside the parentheses.

R^2 + r^2 + R r = 6^2 + 3^2 + (6 \times 3) = 36 + 9 + 18 = 63

.

Computing the effective area scaling factor of the frustum bases.

3

Multiply by

\frac{1}{3}\pi h

to find the total volume.

V = \frac{1}{3} \times \pi \times 4 \times 63 = 4 \times 21 \times \pi = 84\pi\text{ cm}^3

.

Completing the frustum volume calculation.

Key Concept

Volume of a Frustum of a Cone

Question 7733Question →

What is the value of the definite integral $\int_{1}^{2} \left(3x^2 - \frac{4}{x^2}\right) \, dx$ ?

5

6

9

10

Show answer & explanation

Answer: 5

Answer

5

Integrating

3x^2 - 4x^{-2}

gives

x^3 + \frac{4}{x}

. Substituting the upper bound

x=2

gives

8 + 2 = 10

, and substituting the lower bound

x=1

gives

1 + 4 = 5

. The difference is

10 - 5 = 5

.

Step-by-Step Solution

1

Rewrite the integrand with a negative exponent

3x^2 - 4x^{-2}

Converting reciprocal powers into negative exponents simplifies applying the power rule of integration.

2

Find the indefinite integral (antiderivative)

\int (3x^2 - 4x^{-2}) \, dx = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} = x^3 + \frac{4}{x}

Apply the power rule

\int x^n \, dx = \frac{x^{n+1}}{n+1}

to each term.

3

Evaluate the antiderivative at the upper limit

x = 2

2^3 + \frac{4}{2} = 8 + 2 = 10

Substitute the upper limit into the antiderivative expression.

4

Evaluate the antiderivative at the lower limit

x = 1

1^3 + \frac{4}{1} = 1 + 4 = 5

Substitute the lower limit into the antiderivative expression.

5

Subtract the lower limit value from the upper limit value

10 - 5 = 5

According to the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

.

Key Concept

Definite Integration of Polynomial and Power Functions

Question 7734Question →

If $\int (8x^3 - 9\sin(3x) + 2) \, dx = ax^4 + b\cos(3x) + cx + C$ , where $a$ , $b$ , and $c$ are constant coefficients and $C$ is the constant of integration, what is the value of $a + b + c$ ?

Show answer & explanation

Answer: 7

Answer

The value of

a + b + c

is 7.

Integrating term-by-term yields

\int 8x^3 dx = 2x^4

,

\int -9\sin(3x) dx = 3\cos(3x)

, and

\int 2 dx = 2x

. Equating coefficients with

ax^4 + b\cos(3x) + cx

gives

a = 2

,

b = 3

, and

c = 2

. Therefore,

a + b + c = 2 + 3 + 2 = 7

.

Step-by-Step Solution

1

Integrate the polynomial term

8x^3

\int 8x^3 \, dx = 2x^4

, identifying

a = 2

Apply the power rule of integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

-9\sin(3x)

\int -9\sin(3x) \, dx = 3\cos(3x)

, identifying

b = 3

Apply the standard trigonometric integral formula:

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

3

Integrate the constant term

2

\int 2 \, dx = 2x

, identifying

c = 2

The integral of a constant

k

with respect to

x

is

kx

.

4

Calculate the requested sum

a + b + c

a + b + c = 2 + 3 + 2 = 7

Summing the coefficients derived from each term's antiderivative.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 7735Question →

A progressive transverse wave traveling along a taut string is governed by the displacement equation $y(x,t) = 0.05 \sin(200\pi t - 10\pi x)$ , where $x$ and $y$ are measured in meters and $t$ in seconds. Calculate the distance, in meters, traveled by the wave front during the time taken for a single particle on the string to complete $15$ full oscillations.

Show answer & explanation

Answer: 3

Answer

The distance traveled by the wave front during 15 full particle oscillations is

3.0\text{ m}

.

Comparing

y(x,t) = 0.05 \sin(200\pi t - 10\pi x)

to the standard wave equation

y = A \sin(\omega t - kx)

gives

\omega = 200\pi\text{ rad/s}

and

k = 10\pi\text{ rad/m}

. The wave speed

v = \frac{\omega}{k} = 20\text{ m/s}

. The time for one full oscillation is

T = \frac{2\pi}{\omega} = 0.01\text{ s}

, so 15 full oscillations take

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

. The distance traveled by the wave front is

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

. Alternatively, because a wave travels a distance of one wavelength

\lambda = \frac{2\pi}{k} = 0.2\text{ m}

during each period (1 full oscillation), in 15 full oscillations the wave travels

15 \times \lambda = 15 \times 0.2\text{ m} = 3.0\text{ m}

.

Step-by-Step Solution

1

Extract wave parameters from the progressive wave equation

Angular frequency

\omega = 200\pi\text{ rad/s}

and wave number

k = 10\pi\text{ rad/m}

.

Matching the given equation

y = 0.05 \sin(200\pi t - 10\pi x)

with the standard form

y = A \sin(\omega t - kx)

identifies

\omega

and

k

.

2

Calculate the wave propagation velocity

v = 20\text{ m/s}

.

Wave speed is given by the relation

v = \frac{\omega}{k} = \frac{200\pi}{10\pi} = 20\text{ m/s}

.

3

Find the period of oscillation and total elapsed time

Period

T = 0.01\text{ s}

, total time

t = 0.15\text{ s}

.

The period

T = \frac{2\pi}{\omega} = \frac{2\pi}{200\pi} = 0.01\text{ s}

. For 15 complete oscillations,

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

.

4

Compute the total distance traveled by the wave

Distance

d = 3.0\text{ m}

.

Using linear motion at constant wave speed,

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

.

Key Concept

Wave equation parameters, particle oscillation period, and wave propagation distance

Estimated Time:2m 0s

Question 7736Question →

Match each physical scenario involving scalar and vector quantities on the left with its corresponding resultant value or component magnitude on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) The particle's net displacement corresponds to 13 m; (2) The resultant of two equal forces of magnitude F at 60 degrees corresponds to F√3; (3) The vertical component of a 40 N force inclined at 60 degrees to the vertical corresponds to 20 N; (4) The resultant of perpendicular forces of 8 N and 15 N corresponds to 17 N.

Each scenario correctly applies vector algebra: 2D displacement resolution yields a 5-12-13 right triangle; the parallelogram rule for equal forces at 60 degrees produces F√3; resolving a force adjacent to the vertical axis uses cos(60°) to give 20 N; and perpendicular 8 N and 15 N forces synthesize to a 17 N resultant using the Pythagorean theorem.

Step-by-Step Solution

1

Calculate net displacement for Item 1

Net x-component:

10\text{ m} - 5\text{ m} = 5\text{ m}

East. Net y-component:

12\text{ m}

North. Magnitude

R = \sqrt{5^2 + 12^2} = 13\text{ m}

.

Displacements along parallel lines subtract scalar-wise, and perpendicular components combine via the Pythagorean theorem.

2

Determine the resultant of two equal forces at 60 degrees for Item 2

R = \sqrt{F^2 + F^2 + 2(F)(F)\cos(60^\circ)} = \sqrt{2F^2 + 2F^2(0.5)} = \sqrt{3F^2} = F\sqrt{3}

.

Applying the parallelogram law of vector addition.

3

Resolve the force vector along the vertical direction for Item 3

F_{\text{vertical}} = F \cos(\theta_{\text{vertical}}) = 40 \cos(60^\circ) = 40 \times 0.5 = 20\text{ N}

.

The component adjacent to the reference angle uses the cosine function.

4

Compute resultant magnitude of orthogonal forces for Item 4

R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ N}

.

Vectors at right angles sum directly using Pythagorean synthesis.

Key Concept

Vector resolution, component synthesis, and parallelogram law of vector addition

Question 7737Question →

The rate of change of linear momentum per unit cross-sectional area is a derived physical quantity. When resolved into fundamental SI base units, which of the following physical quantities has the exact same SI base unit decomposition?

Surface tension

Energy density

Power

Specific heat capacity

Show answer & explanation

Answer: Energy density

Answer

Energy density

The rate of change of momentum is force (

F

), which has SI base units of

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

. Dividing by area (

\text{m}^2

) gives

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

. Energy density is defined as energy per unit volume, which decomposes to

\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

. Both derived quantities possess identical fundamental SI base unit representations.

Step-by-Step Solution

1

Determine the physical definition of the rate of change of linear momentum per unit area.

By Newton's second law, rate of change of momentum equals force (

F = \frac{\Delta p}{\Delta t}

). Dividing by cross-sectional area (

A

) yields force per unit area (pressure or stress),

\frac{F}{A}

.

Force is defined as the time rate of change of linear momentum.

2

Express force per unit area in terms of fundamental SI base units.

\text{Force} = \text{mass} \times \text{acceleration} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

. Therefore,

\frac{F}{A} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Mass (

\text{kg}

), length (

\text{m}

), and time (

\text{s}

) are fundamental base quantities.

3

Analyze energy density in fundamental SI base units.

\text{Energy density} = \frac{\text{Energy}}{\text{Volume}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Decomposing energy into work (force

\times

distance) and dividing by volume reveals its base unit equivalence.

Key Concept

Decomposition of derived quantities into fundamental SI base units

Estimated Time:1m 30s

Question 7738Question →

In the International System of Units (SI), physical quantities are categorized as either fundamental or derived. Which of the following is a fundamental SI base unit?

Joule

Kelvin

Pascal

Watt

Show answer & explanation

Answer: Kelvin

Answer

Kelvin is the fundamental SI base unit.

The kelvin is the standard SI base unit for measuring thermodynamic temperature and is one of the seven defined fundamental units.

Step-by-Step Solution

1

Identify the seven SI fundamental (base) units

The seven fundamental SI base units are the meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).

Fundamental units are basic units that are independent of one another and cannot be expressed in terms of other units.

2

Compare the given options against the set of fundamental units

Among Joule, Kelvin, Pascal, and Watt, only Kelvin is in the list of fundamental SI base units.

Joule, Pascal, and Watt are all derived units defined by combinations of base units.

Key Concept

Fundamental SI Base Units

Estimated Time:35s

Question 7739Question →

What is the maximum value of the trigonometric function $y = 3\sin x + 2$ ?

Show answer & explanation

Answer: 5

Answer

The maximum value of the function is 5.

The basic sine function

\sin x

reaches a maximum value of

1

. Substituting

\sin x = 1

into

y = 3\sin x + 2

yields

y = 3(1) + 2 = 5

.

Step-by-Step Solution

1

Identify the maximum value of the sine term

The range of

\sin x

is

[-1, 1]

, so its maximum value is

1

.

The sine function oscillates between a minimum of

-1

and a maximum of

1

for all real numbers

x

.

2

Calculate the maximum value of the transformed function

y_{\text{max}} = 3(1) + 2 = 5

.

Multiplying by the positive amplitude coefficient

3

scales the peak to

3

, and adding the vertical shift of

2

raises the peak to

5

.

Key Concept

Maximum and Minimum Values of Trigonometric Functions

Question 7740Question →

Find the set of real values of $x$ that satisfies the inequality $\frac{5 - 2x}{3} \ge x - 5$ .

x \ge 4

x \le 4

x \le -4

x \ge -4

Show answer & explanation

Answer: $x \le 4$

Answer

The set of real values satisfying the inequality is

x \le 4

.

Multiplying through by

3

yields

5 - 2x \ge 3x - 15

. Rearranging terms gives

-5x \ge -20

. Dividing both sides by

-5

requires flipping the inequality sign from

\ge

to

\le

, giving

x \le 4

.

Step-by-Step Solution

1

Multiply both sides of the inequality by 3 to clear the fraction.

5 - 2x \ge 3(x - 5)

Eliminating the denominator simplifies the algebraic expression.

2

Expand the right-hand side and collect terms containing

x

on one side and constants on the other.

5 - 2x \ge 3x - 15 \implies -2x - 3x \ge -15 - 5 \implies -5x \ge -20

Group like terms to isolate the variable

x

.

3

Divide both sides by

-5

and reverse the inequality sign.

x \le \frac{-20}{-5} \implies x \le 4

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Solving linear inequalities involving negative coefficient division

Answer

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