All practice questions

What is the numerical value of the simplified expression $\frac{4}{\sqrt{5} - 1} - \sqrt{5}$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of the expression is 1.

Multiplying the top and bottom of

\frac{4}{\sqrt{5} - 1}

by its conjugate

(\sqrt{5} + 1)

simplifies the fraction to

\frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1

. Subtracting

\sqrt{5}

from

\sqrt{5} + 1

leaves 1.

Step-by-Step Solution

1

Rationalize the denominator of the fractional term

\sqrt{5} + 1

Multiply both numerator and denominator by the conjugate

(\sqrt{5} + 1)

to apply the difference of two squares identity

(a-b)(a+b) = a^2 - b^2

in the denominator.

2

Subtract the remaining surd term

1

Subtract

\sqrt{5}

from

\sqrt{5} + 1

, leaving the integer 1.

Key Concept

Rationalization of Binomial Denominators

Question 4262Question →

The volume flow rate

Q

of a viscous liquid flowing through a pipe of radius

r

under a pressure gradient

\frac{\Delta P}{l}

is modeled by the equation:

Q = k \eta^x r^y \left(\frac{\Delta P}{l}\right)^z

where

\eta

is the coefficient of dynamic viscosity and

k

is a dimensionless constant. What are the values of the exponents

x

,

y

, and

z

respectively?

x = -1, y = 4, z = 1

x = 1, y = 4, z = -1

x = -1, y = 3, z = 1

x = 1, y = 2, z = 1

Show answer & explanation

Answer: $x = -1, y = 4, z = 1$

Answer

x = -1

,

y = 4

, and

z = 1

The dimensional representation of volume flow rate is

[L^3 T^{-1}]

, dynamic viscosity is

[M L^{-1} T^{-1}]

, radius is

[L]

, and pressure gradient is

[M L^{-2} T^{-2}]

. Equating the powers of

M

,

L

, and

T

gives

x + z = 0

,

-x - 2z = -1

, and

-x + y - 2z = 3

. Solving these simultaneously yields

x = -1

,

y = 4

, and

z = 1

.

Step-by-Step Solution

1

Determine the fundamental dimensions of each physical quantity

Volume flow rate

Q = \frac{\text{Volume}}{\text{Time}} = [L^3 T^{-1}]

;
Dynamic viscosity

\eta = [M L^{-1} T^{-1}]

;
Radius

r = [L]

;
Pressure gradient

\frac{\Delta P}{l} = \frac{\text{Pressure}}{\text{Length}} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}]

.

Correct base dimensions are required for dimensional analysis.

2

Set up the dimensional equation by substituting the base dimensions into the formula

[L^3 T^{-1}] = [M L^{-1} T^{-1}]^x [L]^y [M L^{-2} T^{-2}]^z = M^{x+z} L^{-x + y - 2z} T^{-x - 2z}

.

The principle of dimensional homogeneity requires both sides of the equation to have matching exponents for

M

,

L

, and

T

.

3

Equate exponents for

M

,

T

, and

L

to form algebraic equations

For

M

:

x + z = 0 \implies z = -x

For

T

:

-x - 2z = -1

For

L

:

-x + y - 2z = 3

.

This creates a linear system of equations for the exponents

x

,

y

, and

z

.

4

Solve the system of linear equations

Substituting

z = -x

into the

T

equation:

-x - 2(-x) = -1 \implies x = -1

.
Hence

z = -(-1) = 1

.
Substituting

x = -1

and

z = 1

into the

L

equation:

-(-1) + y - 2(1) = 3 \implies 1 + y - 2 = 3 \implies y = 4

.

Yields the unique set of exponents

x = -1, y = 4, z = 1

.

Key Concept

Dimensional Analysis and Homogeneity

Estimated Time:2m 0s

Question 4263Question →

In political science, the mass media is classified as a primary agent of political socialization.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Mass media is a secondary agent of political socialization, whereas the family is the primary agent.

The statement is false. The family is the fundamental primary agent of political socialization because it imparts foundational civic orientations during early childhood development. Mass media, educational institutions, and civic associations are secondary agents.

Step-by-Step Solution

1

Recall the distinction between primary and secondary agents of political socialization.

Primary agents (such as the family) influence political values early in childhood through direct, close personal interactions. Secondary agents (such as mass media, schools, peer groups, and political parties) operate later in life or through broader, non-familial channels.

Classification depends on the developmental timing and nature of interaction.

2

Determine the agent category for mass media.

Mass media informs and shapes public opinion among older children and adults, placing it in the secondary agent category.

The mass media reinforces, modifies, or expands existing beliefs formed earlier by primary agents.

Key Concept

Primary vs. Secondary Agents of Political Socialization

Question 4264Question →

Calculate the area, in square units, of the triangle formed by the straight line $3x + 4y - 24 = 0$ and the coordinate axes.

Show answer & explanation

Answer: 24

Answer

The area of the triangle formed by the line and the coordinate axes is 24 square units.

To find the area of the triangle bounded by a straight line and the coordinate axes, determine the magnitude of the

x

-intercept and

y

-intercept. Setting

y = 0

in

3x + 4y - 24 = 0

gives

x = 8

. Setting

x = 0

gives

y = 6

. The vertices of the right triangle are at

(0,0)

,

(8,0)

, and

(0,6)

. The area is calculated as

\frac{1}{2} \times 8 \times 6 = 24

square units.

Step-by-Step Solution

1

Find the

x

-intercept of the straight line

x = 8

, corresponding to the point

(8, 0)

Setting

y = 0

determines where the line crosses the horizontal axis

2

Find the

y

-intercept of the straight line

y = 6

, corresponding to the point

(0, 6)

Setting

x = 0

determines where the line crosses the vertical axis

3

Compute the area of the right-angled triangle formed with the origin

(0,0)

\text{Area} = \frac{1}{2} \times 8 \times 6 = 24

The coordinate axes are perpendicular, making the triangle right-angled with base length 8 and height 6

Key Concept

Area of a triangle bounded by a straight line and the coordinate axes

Estimated Time:1m 0s

Question 4265Question →

An open rectangular box with a square base of side length $x\text{ cm}$ is to be constructed such that its total surface area is $108\text{ cm}^2$ . What is the maximum volume of the box in $\text{cm}^3$ ?

54\text{ cm}^3

72\text{ cm}^3

108\text{ cm}^3

432\text{ cm}^3

Show answer & explanation

Answer: $108\text{ cm}^3$

Answer

108\text{ cm}^3

For an open box with a square base of side

x

and height

h

, total surface area is

A = x^2 + 4xh = 108\text{ cm}^2

. Solving for

h

gives

h = \frac{108 - x^2}{4x}

. Substituting

h

into the volume formula gives

V(x) = x^2 h = 27x - \frac{1}{4}x^3

. Differentiating gives

\frac{dV}{dx} = 27 - \frac{3}{4}x^2

. Setting

\frac{dV}{dx} = 0

yields

x = 6

. The second derivative

\frac{d^2V}{dx^2} = -\frac{3}{2}x

evaluated at

x = 6

is

-9 < 0

, confirming a local maximum. Evaluating

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

yields the correct maximum volume.

Step-by-Step Solution

1

Set up the surface area equation for the open box and express height

h

in terms of

x

.

Surface area

A = x^2 + 4xh = 108 \implies h = \frac{108 - x^2}{4x}

.

An open box with a square base has 1 base face (

x^2

) and 4 vertical side faces (

xh

).

2

Formulate the volume function

V(x)

in terms of

x

.

V(x) = x^2 h = x^2 \left(\frac{108 - x^2}{4x}\right) = \frac{1}{4}(108x - x^3) = 27x - \frac{1}{4}x^3

.

Substitute the expression for

h

into the volume formula

V = x^2 h

.

3

Find the critical point by differentiating

V(x)

with respect to

x

and setting

\frac{dV}{dx} = 0

.

\frac{dV}{dx} = 27 - \frac{3}{4}x^2 = 0 \implies \frac{3}{4}x^2 = 27 \implies x^2 = 36 \implies x = 6\text{ cm}

.

Stationary points occur where the first derivative of the volume function equals zero.

4

Verify that

x = 6

gives a maximum volume and calculate

V(6)

.

\frac{d^2V}{dx^2} = -\frac{3}{2}x \implies \left.\frac{d^2V}{dx^2}\right|_{x=6} = -9 < 0

(maximum). Volume

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

.

The negative second derivative confirms a maximum turning point.

Key Concept

Optimization of physical quantities using the first and second derivative tests.

Estimated Time:2m 0s

Question 4266Question →

If set $A = \{a, b, c, d, e\}$ and set $B = \{c, d, e, f, g\}$ , what is the number of elements in the set $(A \setminus B) \cup (B \setminus A)$ ?

Show answer & explanation

Answer: 4; four

Answer

The number of elements in

(A \setminus B) \cup (B \setminus A)

is 4.

The set difference

A \setminus B

consists of elements in

A

that are not in

B

, which gives

\{a, b\}

. Similarly,

B \setminus A

consists of elements in

B

that are not in

A

, giving

\{f, g\}

. The union

(A \setminus B) \cup (B \setminus A)

is

\{a, b, f, g\}

, which has a cardinality of 4.

Step-by-Step Solution

1

Find the relative difference

A \setminus B

A \setminus B = \{a, b\}

Remove elements of

B

present in

A

.

2

Find the relative difference

B \setminus A

B \setminus A = \{f, g\}

Remove elements of

A

present in

B

.

3

Take the union of the two set differences

(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}

Combine elements from both set differences.

4

Count the number of elements in the resulting set

4 elements

The set

\{a, b, f, g\}

contains 4 distinct elements.

Key Concept

Symmetric Difference of Two Sets

Estimated Time:45s

Question 4267Question →

A circle is inscribed in an isosceles trapezium $ABCD$ with parallel sides $AB$ and $CD$ . If $AB = 18\text{ cm}$ and $CD = 8\text{ cm}$ , what is the area of the region inside the trapezium that lies outside the circle?

(156 - 144\pi)\text{ cm}^2

(312 - 36\pi)\text{ cm}^2

(156 - 36\pi)\text{ cm}^2

(169 - 36\pi)\text{ cm}^2

Show answer & explanation

Answer: $(156 - 36\pi)\text{ cm}^2$

Answer

(156 - 36\pi)\text{ cm}^2

For an isosceles trapezium with an inscribed circle, the sum of opposite sides must be equal (

AB + CD = AD + BC = 26\text{ cm}

), giving slant side length

13\text{ cm}

. Using Pythagoras, the perpendicular distance (height) is

\sqrt{13^2 - 5^2} = 12\text{ cm}

. The area of the trapezium is

\frac{1}{2}(18 + 8)(12) = 156\text{ cm}^2

. The inscribed circle has radius equal to half the height (

6\text{ cm}

), so its area is

\pi \times 6^2 = 36\pi\text{ cm}^2

. Subtracting the circle area from the trapezium area yields

(156 - 36\pi)\text{ cm}^2

.

Step-by-Step Solution

1

Apply the property of a tangential quadrilateral to find the non-parallel sides

For a quadrilateral with an inscribed circle, the sum of opposite sides is equal:

AB + CD = AD + BC = 18 + 8 = 26\text{ cm}

. Since trapezium

ABCD

is isosceles,

AD = BC = 13\text{ cm}

.

Tangential quadrilaterals have equal sums of opposite side lengths.

2

Calculate the height

h

of the trapezium using the Pythagorean theorem

Dropping vertical altitudes from top vertices

C

and

D

creates right triangles at the base with horizontal leg

\frac{18 - 8}{2} = 5\text{ cm}

. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\text{ cm}

.

The height of the trapezium forms the perpendicular leg of the right-angled side triangle.

3

Find the area of the trapezium and the inscribed circle

Area of trapezium

= \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(18 + 8) \times 12 = 156\text{ cm}^2

. The diameter of the inscribed circle equals the height

h = 12\text{ cm}

, so its radius is

r = 6\text{ cm}

. Area of circle

= \pi r^2 = 36\pi\text{ cm}^2

.

The diameter of a circle inscribed between parallel bases equals the vertical height between those bases.

4

Subtract the area of the circle from the area of the trapezium

Remaining Area

= 156 - 36\pi\text{ cm}^2

.

The region inside the trapezium but outside the circle is the difference between their areas.

Key Concept

Perimeter and Area of Composite Figures and Inscribed Shapes

Question 4268Question →

A small business recorded the number of customer inquiries received per day over six consecutive days as follows: $4$ , $7$ , $8$ , $11$ , $13$ , and $17$ . What is the variance of the daily customer inquiries?

Show answer & explanation

Answer: 18

Answer

The variance of the daily customer inquiries is

18

.

To find the variance of the data set

\{4, 7, 8, 11, 13, 17\}

, first calculate the mean:

\bar{x} = \frac{4+7+8+11+13+17}{6} = 10

. Next, compute the squared deviation of each data point from the mean:

(4-10)^2 = 36

,

(7-10)^2 = 9

,

(8-10)^2 = 4

,

(11-10)^2 = 1

,

(13-10)^2 = 9

, and

(17-10)^2 = 49

. Summing these squared deviations gives

108

. Dividing this total by the number of observations (

6

) yields a variance of

18

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the given data set.

\bar{x} = 10

The mean is required as the central point from which deviations are calculated.

2

Determine the squared deviation of each data value from the mean.

(-6)^2 = 36

,

(-3)^2 = 9

,

(-2)^2 = 4

,

1^2 = 1

,

3^2 = 9

,

7^2 = 49

Variance measures the average squared distance of data points from the mean.

3

Sum all calculated squared deviations.

\sum (x - \bar{x})^2 = 36 + 9 + 4 + 1 + 9 + 49 = 108

This provides the total sum of squares for the data set.

4

Divide the total sum of squares by the number of observations (

n = 6

).

\sigma^2 = \frac{108}{6} = 18

Population variance formula is

\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}

.

Key Concept

Variance of Ungrouped Data

Estimated Time:1m 30s

Question 4269Question →

Given that $F(x) = \int (3x^2 - 2\sin x) \, dx$ and $F(0) = 6$ , what is the value of the constant of integration $C$ ?

Show answer & explanation

Answer: 4

Answer

The value of the constant of integration

C

is

4

.

Integrating

3x^2 - 2\sin x

yields

F(x) = x^3 + 2\cos x + C

. Substituting

x = 0

gives

F(0) = 2(1) + C = 2 + C

. Since

F(0) = 6

, setting

2 + C = 6

yields

C = 4

.

Step-by-Step Solution

1

Integrate the function

f(x) = 3x^2 - 2\sin x

with respect to

x

F(x) = x^3 + 2\cos x + C

The antiderivative of

3x^2

is

x^3

and the antiderivative of

-2\sin x

is

2\cos x

.

2

Evaluate

F(0)

using the antiderivative expression

F(0) = 0^3 + 2\cos(0) + C = 2 + C

Since

\cos(0) = 1

, the term

2\cos(0)

simplifies to

2

.

3

Solve for the integration constant

C

using

F(0) = 6

C = 4

Subtracting

2

from both sides of

2 + C = 6

yields

C = 4

.

Key Concept

Indefinite integration of polynomial and trigonometric functions with initial conditions

Question 4270Question →

A $0.50\text{ kg}$ mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is $0.16\text{ J}$ and the force constant of the spring is $32\text{ N/m}$ . What is the speed of the mass, in $\text{m/s}$ , at the instant when the magnitude of its acceleration is $3.84\text{ m/s}^2$ ?

Show answer & explanation

Answer: 0.64

Answer

The speed of the mass at that instant is

0.64\text{ m/s}

.

Using the relation for total mechanical energy

E = \frac{1}{2}kA^2

, the amplitude is

A = 0.10\text{ m}

. The angular frequency is

\omega = \sqrt{k/m} = 8.0\text{ rad/s}

. From

|a| = \omega^2 |x|

, the displacement magnitude when acceleration is

3.84\text{ m/s}^2

is

|x| = 0.06\text{ m}

. Substituting these values into

v = \omega \sqrt{A^2 - x^2}

yields

v = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the simple harmonic motion

\omega = 8.0\text{ rad/s}

The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.

2

Calculate the amplitude of oscillation from total energy

A = 0.10\text{ m}

The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.

3

Find the magnitude of displacement corresponding to the given acceleration

|x| = 0.06\text{ m}

In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.

4

Calculate the speed at this displacement using the SHM velocity-displacement relation

v = 0.64\text{ m/s}

Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Key Concept

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion

Question 4271Question →

Match each vector scenario on the left with its correct resultant magnitude or value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match perpendicular forces of 6 N and 8 N to 10 N; minimum resultant of 7 N and 12 N forces to 5 N; two 15 N forces at 120 degrees to 15 N; and 3D displacement vector to 13 m.

Each vector calculation correctly applies the geometric or algebraic properties of vectors: orthogonal vector resolution, opposite-direction subtraction, law of cosines for equal magnitudes at 120 degrees, and 3D component magnitude synthesis.

Step-by-Step Solution

1

Calculate the magnitude of perpendicular vectors

R = \sqrt{6^2 + 8^2} = 10\text{ N}

Perpendicular vectors form a right-angled triangle, so the Pythagorean theorem applies.

2

Find the minimum resultant magnitude of two vectors

R_{\text{min}} = 12\text{ N} - 7\text{ N} = 5\text{ N}

Minimum resultant occurs when vectors act collinear in opposite directions.

3

Determine the resultant of two equal vectors at

120^\circ

R = 15\text{ N}

Using the cosine rule

R = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ}

, since

\cos 120^\circ = -0.5

,

R = F

.

4

Compute the magnitude of the 3D displacement vector

|\mathbf{r}| = \sqrt{3^2 + 4^2 + 12^2} = 13\text{ m}

3D magnitude is calculated using the square root of the sum of squared orthogonal components.

Key Concept

Vector Addition, Resolution, and Magnitude Evaluation

Question 4272Question →

A projectile is launched from ground level over flat terrain. At time $t = 2\text{ s}$ after launch, the projectile passes through a point located $60\text{ m}$ horizontally and $60\text{ m}$ vertically from its launch point. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the total horizontal range of the projectile in meters?

Show answer & explanation

Answer: 240

Answer

The total horizontal range of the projectile is

240\text{ m}

.

The horizontal motion occurs at a constant velocity of

30\text{ m/s}

calculated from

\frac{60\text{ m}}{2\text{ s}}

. Substituting the vertical position (

60\text{ m}

) and time (

2\text{ s}

) into

y = u_y t - 5t^2

yields an initial vertical velocity of

40\text{ m/s}

. The total duration in the air is

T = \frac{2(40)}{10} = 8\text{ s}

. The total horizontal range is therefore

30\text{ m/s} \times 8\text{ s} = 240\text{ m}

.

Step-by-Step Solution

1

Determine the horizontal component of velocity

v_x = 30\text{ m/s}

Horizontal velocity remains constant throughout flight because there is no horizontal acceleration.

2

Determine the initial vertical component of velocity

u_y = 40\text{ m/s}

Applying the vertical displacement equation

y = u_y t - \frac{1}{2}g t^2

with

y = 60\text{ m}

,

t = 2\text{ s}

, and

g = 10\text{ m/s}^2

.

3

Calculate the total time of flight

T = 8\text{ s}

The projectile completes its full parabolic trajectory when vertical displacement returns to zero, given by

T = \frac{2 u_y}{g}

.

4

Calculate the total horizontal range

R = 240\text{ m}

The total range is the product of the constant horizontal velocity component and total time of flight (

R = v_x \times T

).

Key Concept

Independence of horizontal and vertical components of projectile motion

Question 4273Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to measure the internal diameter of a beaker. Before making the measurement, the jaws are brought together tightly; the zero mark of the Vernier scale lies to the left of the zero mark of the main scale, and the $6\text{th}$ division on the Vernier scale coincides with a main scale division. If the observed reading for the internal diameter of the beaker is $3.43\text{ cm}$ , what is the actual internal diameter of the beaker?

3.39\text{ cm}

3.43\text{ cm}

3.47\text{ cm}

3.49\text{ cm}

Show answer & explanation

Answer: $3.47\text{ cm}$

Answer

The actual internal diameter of the beaker is

3.47\text{ cm}

.

The instrument possesses a negative zero error because the Vernier zero mark lies to the left of the main scale zero mark when closed. The magnitude of this negative error is calculated as

-(10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}

. Applying the standard correction formula

\text{Actual Value} = \text{Observed Value} - \text{Zero Error}

gives

3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero error =

- (10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}

When the zero mark of the Vernier scale lies to the left of the main scale zero mark, the zero error is negative. For a 10-division Vernier scale, the magnitude is given by

(10 - N) \times \text{least count}

, where

N=6

is the coinciding division.

2

Apply the zero error correction formula

Actual Reading = Observed Reading - Zero Error

The true reading is obtained by subtracting the zero error (with its sign) from the observed value.

3

Calculate the actual internal diameter

Actual Diameter =

3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}

Subtracting a negative zero error is mathematically equivalent to adding its magnitude to the observed reading.

Key Concept

Vernier Caliper Zero Error Correction

Estimated Time:1m 30s

Question 4274Question →

A ship uses a sonar device to determine the depth of the ocean floor. A sound pulse sent vertically downward reflects off the seabed and is detected $0.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at that location?

450\text{ m}

900\text{ m}

1800\text{ m}

2500\text{ m}

Show answer & explanation

Answer: $450\text{ m}$

Answer

450\text{ m}

The correct answer is

450\text{ m}

because sound travels to the ocean floor and back, covering twice the actual depth. Dividing the total distance of

900\text{ m}

by

2

yields the one-way depth of

450\text{ m}

.

Step-by-Step Solution

1

Calculate the total distance traveled by the sound wave

d_{\text{total}} = 1500\text{ m/s} \times 0.6\text{ s} = 900\text{ m}

Total distance traveled by the wave equals speed multiplied by total elapsed time.

2

Determine the one-way depth of the ocean floor

\text{Depth} = \frac{d_{\text{total}}}{2} = \frac{900\text{ m}}{2} = 450\text{ m}

An echo travels down to the seabed and back to the receiver, so the ocean depth is half the total distance covered.

Key Concept

Echo distance calculation

Estimated Time:45s

Question 4275Question →

Five daily rainfall measurements (in mm) recorded in a city are $3, 6, 7, 9,$ and $15$ . What is the standard deviation of these rainfall measurements?

3.2

4

16

20

Show answer & explanation

Answer: $4$

Answer

The standard deviation of the rainfall measurements is

4

.

First compute the mean of the data set:

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

. Next, sum the squared deviations from the mean:

(3-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2 + (15-8)^2 = 25 + 4 + 1 + 1 + 49 = 80

. Dividing by the number of observations (

5

) yields the variance

\sigma^2 = \frac{80}{5} = 16

. Taking the square root gives the standard deviation

\sigma = \sqrt{16} = 4

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the dataset

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = \frac{40}{5} = 8

The mean is needed to find deviations for each data point.

2

Find the deviation of each number from the mean and square it

(3-8)^2 = 25

,

(6-8)^2 = 4

,

(7-8)^2 = 1

,

(9-8)^2 = 1

,

(15-8)^2 = 49

Squaring ensures all deviation values are non-negative.

3

Sum the squared deviations and divide by the total number of items (

n = 5

) to find the variance

\text{Variance } (\sigma^2) = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16

Variance measures the average squared distance from the mean.

4

Take the square root of the variance to obtain standard deviation

\text{Standard Deviation } (\sigma) = \sqrt{16} = 4

Standard deviation returns the dispersion measure back to the original units.

Key Concept

Standard Deviation of Ungrouped Data

Question 4276Question →

A train accelerates uniformly along a straight track from an initial velocity of $10\text{ m/s}$ to a final velocity of $30\text{ m/s}$ over a distance of $100\text{ m}$ . What is the magnitude of the acceleration of the train?

0.2\text{ m/s}^2

4\text{ m/s}^2

5\text{ m/s}^2

8\text{ m/s}^2

Show answer & explanation

Answer: $4\text{ m/s}^2$

Answer

4\text{ m/s}^2

Using the equation of motion

v^2 = u^2 + 2as

, substitute

u = 10\text{ m/s}

,

v = 30\text{ m/s}

, and

s = 100\text{ m}

. This gives

30^2 = 10^2 + 2(a)(100)

, which simplifies to

900 - 100 = 200a

, or

800 = 200a

. Solving for acceleration yields

a = 4\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the given kinematic values from the problem statement.

Initial velocity

u = 10\text{ m/s}

, final velocity

v = 30\text{ m/s}

, and displacement

s = 100\text{ m}

.

Listing known values helps in selecting the appropriate equation of motion.

2

Select the linear motion formula relating

u

,

v

,

s

, and acceleration

a

.

v^2 = u^2 + 2as

This formula connects initial velocity, final velocity, distance, and acceleration without requiring time

t

.

3

Substitute the values into the equation and solve for

a

.

30^2 = 10^2 + 2(a)(100) \implies 900 = 100 + 200a \implies 800 = 200a \implies a = 4\text{ m/s}^2

Algebraic rearrangement yields the magnitude of acceleration.

Key Concept

Equations of Uniformly Accelerated Motion

Estimated Time:45s

Question 4277Question →

A water pump driven by an engine with an efficiency of $80\%$ raises water from an underground tank of depth $20\text{ m}$ and discharges it through a nozzle of cross-sectional area $10\text{ cm}^2$ at a steady speed of $10\text{ m/s}$ . Taking the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the minimum input power rating (in $\text{W}$ ) required for the engine?

Show answer & explanation

Answer: 3125

Answer

The minimum input power rating required for the engine is

3125\text{ W}

.

The engine must supply power to lift

10\text{ kg}

of water per second through a vertical height of

20\text{ m}

while accelerating it to

10\text{ m/s}

. The useful power output is

2000\text{ W}

(potential)

+ 500\text{ W}

(kinetic)

= 2500\text{ W}

. Accounting for an engine efficiency of

80\%

, the total input power is

\frac{2500}{0.80} = 3125\text{ W}

.

Step-by-Step Solution

1

Determine the mass of water discharged per unit time (mass flow rate).

\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}

Water is moving through a cross-sectional area at a constant velocity.

2

Calculate the useful output power required to lift the water and impart kinetic energy.

P_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}

The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.

3

Calculate the total input power using engine efficiency.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}

Efficiency is the ratio of useful power output to total power input.

Key Concept

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations

Question 4278Question →

What is the remainder when the polynomial $P(x) = x^3 + 3x^2 - 2x + 4$ is divided by $x - 1$ ?

6

8

10

4

Show answer & explanation

Answer: $6$

Answer

The remainder when

P(x)

is divided by

x - 1

is

6

.

According to the Remainder Theorem, dividing a polynomial

P(x)

by a linear divisor

x - a

leaves a remainder equal to

P(a)

. For the divisor

x - 1

, setting

x - 1 = 0

yields

x = 1

. Substituting

x = 1

into

P(x) = x^3 + 3x^2 - 2x + 4

gives

1 + 3 - 2 + 4 = 6

. Therefore, the value

6

is the correct remainder.

Step-by-Step Solution

1

Apply the Remainder Theorem

To find the remainder when

P(x)

is divided by

x - a

, set

x - 1 = 0

, giving

x = 1

. The remainder is equal to

P(1)

.

By the Remainder Theorem, dividing a polynomial

P(x)

by

(x - a)

yields a remainder of

P(a)

.

2

Substitute

x = 1

into

P(x) = x^3 + 3x^2 - 2x + 4

P(1) = (1)^3 + 3(1)^2 - 2(1) + 4 = 1 + 3 - 2 + 4 = 6

.

Direct evaluation of the expression at

x = 1

yields the numerical value of the remainder.

Key Concept

The Remainder Theorem states that when a polynomial

P(x)

is divided by a linear factor

(x - a)

, the remainder is

P(a)

.

Question 4279Question →

Examine the poetic extract below and complete the literary analysis statement by identifying the correct figure of speech.

Fill in the blanks below

"The western wave was all a-flame,
The day was well nigh done!"
— Samuel Taylor Coleridge, The Rime of the Ancient Mariner

In the excerpt above, the word 'wave' is employed to represent the entire ocean. The figure of speech in which a part of an entity is used to stand for the whole is known as .

Show answer & explanation

Answer

synecdoche

Synecdoche is a figure of speech where a part of something represents the whole. In Coleridge's verse, 'wave' (an individual component of the body of water) is used to signify the entire ocean.

Step-by-Step Solution

1

Analyze the target word 'wave' in relation to the surrounding context of the poem.

The word 'wave' is used by the poet to denote the entire sea or ocean.

A wave is a constituent physical part of the sea.

2

Classify the specific trope that uses a part-for-whole substitution.

Synecdoche is defined as substituting a part for the whole (or the whole for a part).

Because 'wave' (a part) represents the ocean (the whole), the literary device is synecdoche.

Key Concept

Synecdoche vs. Metonymy in Poetic Imagery

Estimated Time:2m 0s

Question 4280Question →

Match each physical wave description on the left with its definitive wave classification on the right based on propagation mechanisms and oscillation directions.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Electromagnetic field oscillations without a medium match with Non-mechanical transverse progressive wave; Periodic compressions/rarefactions parallel to wave motion match with Longitudinal mechanical progressive wave; Superposition of opposing identical waves with zero net energy flow matches with Stationary (standing) mechanical wave; Perpendicular particle vibrations in an elastic medium match with Transverse mechanical progressive wave.

Each wave type is categorized by three primary criteria: medium requirement (mechanical requires a medium, non-mechanical does not), vibration alignment (transverse is perpendicular, longitudinal is parallel), and energy movement (progressive transfers energy continuously, stationary confines energy between nodes). Oscillating fields in vacuum represent non-mechanical transverse waves; compressions/rarefactions represent longitudinal mechanical waves; opposing wave superposition forms stationary waves; and perpendicular elastic vibrations represent transverse mechanical waves.

Step-by-Step Solution

1

Classify the wave mechanism based on medium requirement.

Waves requiring an elastic medium (particle vibrations) are mechanical; waves capable of propagating through a vacuum via field oscillations are non-mechanical.

Determines whether the wave is mechanical or non-mechanical.

2

Classify the direction of particle or field oscillation relative to propagation direction.

Oscillations parallel to wave motion form longitudinal waves; oscillations perpendicular to wave motion form transverse waves.

Distinguishes between longitudinal and transverse wave modes.

3

Determine energy transfer characteristics (progressive vs. stationary).

Continuous outward transfer of energy indicates a progressive wave, whereas trapped energy bounded by fixed nodes/antinodes resulting from wave superposition indicates a standing (stationary) wave.

Differentiates traveling waves from standing wave patterns.

4

Map each wave description to its complete wave classification.

All four wave descriptions are uniquely matched with their corresponding taxonomy classifications.

Completes the matching process.

Key Concept

Classification of waves by medium requirement, oscillation direction, and energy propagation mode

Estimated Time:1m 30s

Answer

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