Tüm alıştırma soruları

Complete the sentence below by supplying the most appropriate prepositions to fill the blanks.

Aşağıdaki boşlukları doldurun

Although the chief financial officer claimed to be fully conversant the new auditing regulations, her report was found to be completely at variance the statutory guidelines.

Cevabı ve açıklamayı göster

Cevap

The first blank is filled by 'with' and the second blank is filled by 'with'.

The adjective 'conversant' correctly pairs with the dependent preposition 'with' when indicating familiarity with rules or facts. Similarly, the fixed idiomatic expression 'at variance' takes the preposition 'with' to signify inconsistency or opposition.

Adım Adım Çözüm

1

Identify the dependent preposition required by the adjective 'conversant'.

The adjective 'conversant' standardly collocates with the preposition 'with' (meaning familiar with or knowledgeable about).

English grammatical convention dictates that one is conversant with a subject.

2

Identify the fixed prepositional phrase containing 'at variance'.

The idiom 'at variance' requires the preposition 'with' to show conflict or disagreement with a set of standards.

The standard prepositional expression is 'at variance with'.

Anahtar Kavram

Dependent Prepositions and Fixed Prepositional Phrases

Tahmini Süre:1m 0s

Soru 3822Soru →

If $y = \ln(\cos 2x)$ , what is $\frac{dy}{dx}$ ?

-2\tan 2x

2\tan 2x

-\tan 2x

-2\cot 2x

Cevabı ve açıklamayı göster

Cevap: $-2\tan 2x$

Cevap

-2\tan 2x

Applying the chain rule to

y = \ln(\cos 2x)

yields

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2\frac{\sin 2x}{\cos 2x} = -2\tan 2x

.

Adım Adım Çözüm

1

Identify the inner and outer functions for the chain rule.

Let

u = \cos 2x

, so

y = \ln u

.

The given function

y = \ln(\cos 2x)

is a composite transcendental function.

2

Differentiate

y

with respect to

u

, and

u

with respect to

x

.

\frac{dy}{du} = \frac{1}{u} = \frac{1}{\cos 2x}

, and

\frac{du}{dx} = -2\sin 2x

.

The derivative of

\ln u

is

\frac{1}{u}

and the derivative of

\cos 2x

is

-2\sin 2x

using the chain rule.

3

Apply the chain rule

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

and simplify.

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-2\sin 2x) = -2\left(\frac{\sin 2x}{\cos 2x}\right) = -2\tan 2x

.

Using the trigonometric identity

\frac{\sin 2x}{\cos 2x} = \tan 2x

simplifies the expression into standard form.

Anahtar Kavram

Differentiation of composite logarithmic and trigonometric functions using the Chain Rule

Tahmini Süre:1m 0s

Soru 3823Soru →

Which prepositions correctly complete the blanks in the sentence below?

Aşağıdaki boşlukları doldurun

The union leaders argued that the new management policy was completely incompatible existing labor laws, leaving the workforce entirely indifferent any proposed compromises.

Cevabı ve açıklamayı göster

Cevap

The correct prepositions are 'with' for the first blank and 'to' for the second blank.

The adjective 'incompatible' requires the dependent preposition 'with' to signal inconsistency, while 'indifferent' requires 'to' to indicate apathy or unconcern.

Adım Adım Çözüm

1

Identify the dependent preposition required by the adjective 'incompatible'.

The adjective 'incompatible' collocates strictly with 'with' when indicating conflict or inability to coexist.

Standard English usage rules require 'incompatible with'.

2

Identify the dependent preposition required by the adjective 'indifferent'.

The adjective 'indifferent' collocates strictly with 'to' when indicating lack of interest or concern.

Standard grammatical collocation dictates that one is 'indifferent to' something.

Anahtar Kavram

Dependent Prepositions following Adjectives

Tahmini Süre:1m 0s

Soru 3824Soru →

A car starts from rest and accelerates uniformly to a velocity of $16\text{ m/s}$ in $4\text{ s}$ . It continues at this constant velocity for $8\text{ s}$ , and then decelerates uniformly to rest in another $4\text{ s}$ . What is the average speed of the car for the entire journey?

12\text{ m/s}

16\text{ m/s}

8\text{ m/s}

24\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $12\text{ m/s}$

Cevap

12\text{ m/s}

The average speed of an object undergoing multi-stage motion is defined as the total distance traveled divided by the total time taken. The total distance covered is

32\text{ m}

(during acceleration)

+ 128\text{ m}

(during constant velocity)

+ 32\text{ m}

(during deceleration)

= 192\text{ m}

. Dividing

192\text{ m}

by the total time of

16\text{ s}

yields

12\text{ m/s}

.

Adım Adım Çözüm

1

Calculate the distance covered in each stage of motion.

Acceleration stage (

s_1

):

\frac{0 + 16}{2} \times 4 = 32\text{ m}

. Constant velocity stage (

s_2

):

16 \times 8 = 128\text{ m}

. Deceleration stage (

s_3

):

\frac{16 + 0}{2} \times 4 = 32\text{ m}

.

The total distance is the sum of distances traveled during acceleration, constant speed, and deceleration.

2

Find total distance traveled and total time taken.

Total distance (

S

) =

32\text{ m} + 128\text{ m} + 32\text{ m} = 192\text{ m}

. Total time (

T

) =

4\text{ s} + 8\text{ s} + 4\text{ s} = 16\text{ s}

.

Average speed requires total distance divided by total elapsed time.

3

Compute the average speed.

Average speed =

\frac{S}{T} = \frac{192\text{ m}}{16\text{ s}} = 12\text{ m/s}

.

Dividing total distance by total time gives the average speed over the entire motion.

Anahtar Kavram

Average Speed in Multi-Stage Motion

Tahmini Süre:1m 30s

Soru 3825Soru →

If $\log_4 x + \log_2 x = 6$ , find the value of $x$ .

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The value of

x

is 16.

Using the change of base identity

\log_b a = \frac{\log_c a}{\log_c b}

, we convert

\log_4 x

into

\frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

. Substituting this into the equation yields

\frac{3}{2} \log_2 x = 6

, which simplifies to

\log_2 x = 4

, leading directly to

x = 2^4 = 16

.

Adım Adım Çözüm

1

Apply the change of base formula to

\log_4 x

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

Converting all terms to a common base (base 2) simplifies addition of logarithmic terms.

2

Substitute the expression back into the original equation and collect like terms

\frac{1}{2}\log_2 x + \log_2 x = \frac{3}{2}\log_2 x = 6

Adding the coefficients of

\log_2 x

gives

\frac{3}{2}

.

3

Isolate

\log_2 x

\log_2 x = 6 \cdot \frac{2}{3} = 4

Multiplying both sides by

\frac{2}{3}

isolates the logarithmic term.

4

Convert from logarithmic form to exponential form

x = 2^4 = 16

If

\log_b a = c

, then

a = b^c

.

Anahtar Kavram

Change of base rule for logarithms

Soru 3826Soru →

What is the positive integer solution to the logarithmic equation $\log_2 x - 3\log_x 2 = 2$ ?

8

6

3

12

Cevabı ve açıklamayı göster

Cevap: $8$

Cevap

The positive integer solution is

8

.

Using the change of base identity

\log_x 2 = \frac{1}{\log_2 x}

, we rewrite the equation as

u - \frac{3}{u} = 2

where

u = \log_2 x

. Rearranging yields the quadratic

u^2 - 2u - 3 = 0

, which factors as

(u - 3)(u + 1) = 0

. This gives

u = 3

or

u = -1

. Converting back to

x = 2^u

, we get

x = 2^3 = 8

or

x = 2^{-1} = \frac{1}{2}

. Since the question asks for the positive integer solution, the correct answer is 8.

Adım Adım Çözüm

1

Apply the change of base rule to express

\log_x 2

in terms of base 2.

\log_2 x - \frac{3}{\log_2 x} = 2

The change of base identity states that

\log_b a = \frac{1}{\log_a b}

.

2

Substitute

u = \log_2 x

into the equation and clear the fraction.

u - \frac{3}{u} = 2 \implies u^2 - 2u - 3 = 0

Multiplying through by

u

(where

u \neq 0

) transforms the equation into standard quadratic form.

3

Factor the quadratic equation to solve for

u

.

(u - 3)(u + 1) = 0 \implies u = 3 \text{ or } u = -1

Finding the roots of the quadratic equation in terms of

u

.

4

Convert back to

x

using

x = 2^u

and select the positive integer solution.

For

u = 3

:

x = 2^3 = 8

. For

u = -1

:

x = 2^{-1} = \frac{1}{2}

.

The question asks specifically for the positive integer solution, which is 8.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction

Soru 3827Soru →

If $y = 2e^{4x} + \ln(2x + 1)$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

Differentiating

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

yields

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

. Evaluating this derivative at

x = 0

gives

8(e^0) + \frac{2}{2(0) + 1} = 8(1) + 2 = 10

.

Adım Adım Çözüm

1

Differentiate each term of

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

By the chain rule,

\frac{d}{dx}(ae^{kx}) = ak e^{kx}

and

\frac{d}{dx}(\ln(u(x))) = \frac{u'(x)}{u(x)}

.

2

Evaluate the derivative at

x = 0

\left.\frac{dy}{dx}\right|_{x=0} = 8e^{0} + \frac{2}{2(0) + 1} = 8(1) + \frac{2}{1} = 10

Substitute

x = 0

into the derived expression and simplify using

e^0 = 1

.

Anahtar Kavram

Differentiation of exponential and logarithmic functions using the chain rule

Soru 3828Soru →

Arrange the following geographic regions in Nigeria in descending order of their average population density, from the highest population density to the lowest population density.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct descending order of population density is: Anambra-Imo Plateau, Kano Close-Settled Zone, Niger River Valley in the Middle Belt, and Borgu Plains of the Kainji Basin.

Nigeria's population density follows a distinct spatial structure. The Anambra-Imo Plateau in the South-East is the most densely settled region due to heavy commercialization and historical population growth. The Kano Close-Settled Zone in the Sudan Savanna represents a major northern population core owing to fertile hydromorphic soils and intensive trade. The Niger River Valley in the Middle Belt is characterized by low population density due to historical depopulation and vector-borne diseases. The Borgu Plains represent one of the least densely populated zones in Nigeria due to game reserves and poor agricultural potential.

Adım Adım Çözüm

1

Identify the region with the highest population density in Nigeria

The Anambra-Imo Plateau in the South-East is identified as the highest density zone (>500 persons/km²).

High agricultural land pressure and commercial hub concentration make the South-East one of the most densely populated areas in West Africa.

2

Identify the prominent northern high-density agricultural zone

The Kano Close-Settled Zone ranks second in density.

Intensive peri-urban agriculture and historical commercial status create a major population cluster in the Sudan Savanna.

3

Identify the sparsely populated Middle Belt zone

The Niger River Valley ranks third, exhibiting low population density.

Physical and historical constraints such as tsetse flies, seasonal inundation, and past slave raiding kept population levels low.

4

Identify the peripheral area with the lowest population density

The Borgu Plains rank lowest in population density.

Poor agricultural soils, forest conservation areas, and remote location contribute to very low human settlement density.

Anahtar Kavram

Spatial variation of population density and environmental/historical settlement factors in Nigeria

Soru 3829Soru →

If $y = e^{2x} \tan x$ , what is $\frac{dy}{dx}$ ?

e^{2x}(2\tan x + \sec^2 x)

e^{2x}(\tan x + \sec^2 x)

e^{2x}(2\tan x - \sec^2 x)

2e^{2x} \sec^2 x

Cevabı ve açıklamayı göster

Cevap: $e^{2x}(2\tan x + \sec^2 x)$

Cevap

e^{2x}(2\tan x + \sec^2 x)

Applying the product rule

\frac{d}{dx}[uv] = u'v + uv'

with

u = e^{2x}

and

v = \tan x

gives

u' = 2e^{2x}

and

v' = \sec^2 x

. Substituting these into the formula yields

2e^{2x}\tan x + e^{2x}\sec^2 x = e^{2x}(2\tan x + \sec^2 x)

.

Adım Adım Çözüm

1

Identify the component functions for the product rule

Let

u(x) = e^{2x}

and

v(x) = \tan x

.

The given function

y = e^{2x} \tan x

is a product of two functions.

2

Differentiate each component function separately

\frac{du}{dx} = 2e^{2x}

by the chain rule, and

\frac{dv}{dx} = \sec^2 x

.

The derivative of

e^{kx}

is

k e^{kx}

and the derivative of

\tan x

is

\sec^2 x

.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

\frac{dy}{dx} = e^{2x} \cdot \sec^2 x + \tan x \cdot 2e^{2x}

.

Combining the terms according to the standard product rule.

4

Factor out the common term

e^{2x}

\frac{dy}{dx} = e^{2x}(2\tan x + \sec^2 x)

.

Simplifying to match standard exam response format.

Anahtar Kavram

Differentiation of Exponential and Trigonometric Functions using the Product Rule

Tahmini Süre:1m 30s

Soru 3830Soru →

A longitudinal mechanical wave travels through Medium X with a propagation speed of $340\text{ m/s}$ . The distance between two consecutive compressions in Medium X is $0.68\text{ m}$ . When the wave propagates across a boundary into Medium Y, the distance between a compression and the immediately adjacent rarefaction is measured as $1.70\text{ m}$ . What is the speed of propagation of the wave in Medium Y?

1700\text{ m/s}

850\text{ m/s}

340\text{ m/s}

680\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $1700\text{ m/s}$

Cevap

The speed of propagation of the wave in Medium Y is

1700\text{ m/s}

.

In Medium X, the distance between consecutive compressions gives a full wavelength of

0.68\text{ m}

, yielding a wave frequency of

f = \frac{340\text{ m/s}}{0.68\text{ m}} = 500\text{ Hz}

. Because frequency depends solely on the source, it remains

500\text{ Hz}

in Medium Y. In Medium Y, the distance between a compression and the adjacent rarefaction is half a wavelength, making

\lambda_Y = 2 \times 1.70\text{ m} = 3.40\text{ m}

. Multiplying frequency by the new wavelength gives the speed in Medium Y as

v_Y = 500\text{ Hz} \times 3.40\text{ m} = 1700\text{ m/s}

.

Adım Adım Çözüm

1

Determine the frequency of the wave using parameters from Medium X.

In longitudinal waves, the distance between consecutive compressions equals one wavelength, so

\lambda_X = 0.68\text{ m}

. Using

v = f \cdot \lambda

, frequency

f = \frac{340\text{ m/s}}{0.68\text{ m}} = 500\text{ Hz}

.

Frequency is determined by the wave source and remains unchanged when a wave transitions between different media.

2

Calculate the wavelength of the wave in Medium Y.

The distance from a compression to the immediately adjacent rarefaction is half of a wavelength (

\frac{\lambda_Y}{2} = 1.70\text{ m}

). Therefore,

\lambda_Y = 2 \times 1.70\text{ m} = 3.40\text{ m}

.

A full wavelength spans from compression to compression or rarefaction to rarefaction.

3

Calculate the wave propagation speed in Medium Y.

v_Y = f \cdot \lambda_Y = 500\text{ Hz} \times 3.40\text{ m} = 1700\text{ m/s}

.

Applying the wave speed formula with constant frequency.

Anahtar Kavram

Wave Propagation across Boundaries and Longitudinal Wave Characteristics

Tahmini Süre:2m 0s

Soru 3831Soru →

Find the value of $x$ that satisfies the exponential equation $27^{x - 1} = \frac{9^{x + 1}}{3^{x - 5}}$ .

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of

x

is 5.

Rewriting the terms with a common base of 3 transforms the equation into

3^{3x-3} = 3^{x+7}

. Equating exponents gives

3x - 3 = x + 7

, which yields

x = 5

.

Adım Adım Çözüm

1

Convert all terms to base 3

27^{x-1} = 3^{3x-3}

and

9^{x+1} = 3^{2x+2}

Laws of indices require identical bases to manipulate exponents.

2

Apply division rule of indices to the right-hand side

\frac{3^{2x+2}}{3^{x-5}} = 3^{x+7}

When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Equate the exponents and solve for

x

3x - 3 = x + 7 \implies 2x = 10 \implies x = 5

If

a^f(x) = a^g(x)

for

a > 0

and

a \neq 1

, then

f(x) = g(x)

.

Anahtar Kavram

Solving exponential equations using base reduction and exponent laws

Tahmini Süre:1m 30s

Soru 3832Soru →

Complete the sentence below by supplying the most appropriate prepositions to fill the blanks.

Aşağıdaki boşlukları doldurun

The managing director remained impervious public criticism, maintaining that the restructuring plan was ultimately conducive the company's long-term growth.

Cevabı ve açıklamayı göster

Cevap

The correct preposition for both blanks is 'to'.

The word 'impervious' forms a fixed prepositional collocation with 'to', meaning unaffected by something. Similarly, 'conducive' requires the dependent preposition 'to' when indicating contributing toward a result.

Adım Adım Çözüm

1

Determine the correct dependent preposition for the adjective 'impervious'.

The adjective 'impervious' takes the preposition 'to' (impervious to).

In standard English usage, 'impervious' signifies being incapable of being affected, influenced, or damaged, and standardly collocates with 'to'.

2

Determine the correct dependent preposition for the adjective 'conducive'.

The adjective 'conducive' takes the preposition 'to' (conducive to).

The adjective 'conducive' means making a certain outcome likely or possible, and strictly requires the preposition 'to'.

Anahtar Kavram

Adjective-Preposition Collocations

Soru 3833Soru →

A curve is given by the equation $y = \frac{x}{x^2 + 4}$ . What is the maximum value of $y$ on this curve?

\frac{1}{4}

-\frac{1}{4}

2

0

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{4}$

Cevap

The maximum value of

y

on the curve is

\frac{1}{4}

.

To find the maximum value of

y = \frac{x}{x^2 + 4}

, differentiate

y

using the quotient rule to obtain

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting the numerator to zero gives

x = \pm 2

. Since

\frac{dy}{dx}

is positive for

x < 2

and negative for

x > 2

,

x = 2

corresponds to a local maximum. Substituting

x = 2

into

y = \frac{x}{x^2 + 4}

yields

y = \frac{2}{8} = \frac{1}{4}

.

Adım Adım Çözüm

1

Differentiate

y = \frac{x}{x^2 + 4}

using the quotient rule

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative

\frac{dy}{dx}

equals zero.

2

Find the stationary points by setting

\frac{dy}{dx} = 0

4 - x^2 = 0 \implies x = 2 \text{ or } x = -2

A fraction equals zero when its numerator is zero.

3

Determine the nature of the stationary point at

x = 2

\text{For } x < 2, \frac{dy}{dx} > 0; \text{ for } x > 2, \frac{dy}{dx} < 0 \implies x = 2 \text{ is a maximum point}

The derivative changes sign from positive to negative across a maximum point.

4

Evaluate

y

at

x = 2

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

Substituting the

x

-coordinate into the curve equation yields the maximum value of

y

.

Anahtar Kavram

Stationary Points, Maxima, and Minima

Soru 3834Soru →

A curve is defined by the equation $y = x^3 - 4x + 3$ . What is the $y$ -intercept of the tangent line to the curve at the point where $x = 2$ ?

Cevabı ve açıklamayı göster

Cevap: -13

Cevap

The y-intercept of the tangent line is

-13

.

At

x = 2

, substituting into

y = x^3 - 4x + 3

yields

y = 3

. The derivative

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

evaluated at

x = 2

gives

m = 8

. The tangent line equation is

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

. Setting

x = 0

gives the

y

-intercept of

-13

.

Adım Adım Çözüm

1

Calculate the y-coordinate of the point on the curve at

x = 2

y = 3

, giving the point

(2, 3)

The point of contact must lie on the curve.

2

Differentiate the curve equation to find the gradient function

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

The derivative of a function gives the gradient of the tangent at any point

x

.

3

Evaluate the derivative at

x = 2

to find the gradient of the tangent

m

m = 8

Substituting

x = 2

into

3x^2 - 4

yields

3(4) - 4 = 8

.

4

Formulate the linear equation of the tangent line using point

(2, 3)

and gradient

m = 8

y = 8x - 13

Applying

y - y_1 = m(x - x_1)

gives

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

.

5

Extract the y-intercept of the tangent line

y-intercept =

-13

In slope-intercept form

y = mx + c

, the constant term

c = -13

is the y-intercept.

Anahtar Kavram

Tangents to Curves and Line Intercepts

Soru 3835Soru →

A ball $P$ is dropped from rest from the top of a cliff of height $100\text{ m}$ . At the same instant, another ball $Q$ is projected vertically upwards from the base of the cliff along the same vertical line with an initial speed of $50\text{ m/s}$ . Taking $g = 10\text{ m/s}^2$ , at what height above the ground do the two balls meet?

20\text{ m}

50\text{ m}

80\text{ m}

100\text{ m}

Cevabı ve açıklamayı göster

Cevap: $80\text{ m}$

Cevap

The two balls meet at a height of

80\text{ m}

above the ground.

Equating the position equations of both objects gives

100 - 5t^2 = 50t - 5t^2

, which simplifies to

50t = 100

, so

t = 2\text{ s}

. Substituting

t = 2\text{ s}

into the vertical height formula gives

h = 80\text{ m}

above the ground.

Adım Adım Çözüm

1

Set up equations of motion for both balls in terms of time

t

.

For ball

P

(dropped from top): height above ground

h_P = 100 - \frac{1}{2}gt^2 = 100 - 5t^2

.
For ball

Q

(projected from ground): height above ground

h_Q = ut - \frac{1}{2}gt^2 = 50t - 5t^2

.

Both balls move simultaneously under gravity along the same vertical line.

2

Equate the heights

h_P = h_Q

to find the time of meeting

t

.

100 - 5t^2 = 50t - 5t^2 \implies 50t = 100 \implies t = 2\text{ s}

.

The balls pass each other when their heights above the ground are equal.

3

Calculate the height above ground using

t = 2\text{ s}

.

h = 50(2) - 5(2)^2 = 100 - 20 = 80\text{ m}

.

Substituting

t = 2\text{ s}

into either height expression yields the position where they meet.

Anahtar Kavram

Relative vertical motion under gravity

Alternatif Yöntem

Using relative velocity: The relative acceleration between the two balls is

g - g = 0\text{ m/s}^2

. The relative speed of approach is constant at

50\text{ m/s}

. The initial separation is

100\text{ m}

, so time to meet is

t = \frac{100}{50} = 2\text{ s}

. Height above ground is then

h = 100 - \frac{1}{2}(10)(2)^2 = 80\text{ m}

.

Tahmini Süre:1m 30s

Soru 3836Soru →

A solid brass cube with an edge length of $10\text{ cm}$ at $15^\circ\text{C}$ is heated to a temperature of $115^\circ\text{C}$ . If the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the volume of the cube in $\text{cm}^3$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The increase in the volume of the brass cube is

6.0\text{ cm}^3

.

The volume expansion of a solid is given by

\Delta V = V_1 \gamma \Delta T

. The initial volume of the cube is

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

and the temperature change is

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

. Because the expansion occurs in three dimensions, the volume expansivity is

\gamma = 3\alpha = 3 \times 2.0 \times 10^{-5} = 6.0 \times 10^{-5}\text{ K}^{-1}

. Substituting these values yields

\Delta V = 1000 \times 6.0 \times 10^{-5} \times 100 = 6.0\text{ cm}^3

.

Adım Adım Çözüm

1

Calculate the initial volume of the cube

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

The volume of a cube is calculated using

V = L^3

, where

L

is the edge length.

2

Determine the change in temperature

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

The change in temperature is the difference between the final and initial temperatures.

3

Calculate the volume expansivity (cubic expansivity)

\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}

Volume expansivity

\gamma

is three times the linear expansivity

\alpha

for isotropic solids.

4

Compute the increase in volume

\Delta V = V_1 \gamma \Delta T = 1000 \times (6.0 \times 10^{-5}) \times 100 = 6.0\text{ cm}^3

The formula for volume expansion is

\Delta V = V_1 \gamma \Delta T

.

Anahtar Kavram

Thermal expansion of solids: volume expansion (

\Delta V = V_1 \gamma \Delta T

) and the relationship between linear and volume expansivity (

\gamma = 3\alpha

).

Soru 3837Soru →

What is the square root of the surd expression $14 + 6\sqrt{5}$ ?

3 + \sqrt{5}

3 - \sqrt{5}

\sqrt{14} + \sqrt{5}

9 + \sqrt{5}

Cevabı ve açıklamayı göster

Cevap: $3 + \sqrt{5}$

Cevap

3 + \sqrt{5}

Expanding the square of

3 + \sqrt{5}

yields

(3)^2 + 2(3)(\sqrt{5}) + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}

, which accurately equals the original expression under the radical.

Adım Adım Çözüm

1

Set up the general form for the square root of a binomial surd

Let

\sqrt{14 + 6\sqrt{5}} = \sqrt{a} + \sqrt{b}

The square root of a compound surd expression takes the form of a sum of radical terms

2

Square both sides of the equation

14 + 6\sqrt{5} = a + b + 2\sqrt{ab}

Eliminate the outer radical to equate real and surd parts

3

Equate the rational parts and the surd parts

a + b = 14

and

2\sqrt{ab} = 6\sqrt{5} \implies \sqrt{ab} = 3\sqrt{5} = \sqrt{45} \implies ab = 45

Match integer terms together and radical terms together

4

Solve for values of

a

and

b

Two positive numbers with sum

14

and product

45

are

9

and

5

, so

a = 9

and

b = 5

Determine the factors satisfying both equations

5

Substitute

a

and

b

into the radical expression

\sqrt{14 + 6\sqrt{5}} = \sqrt{9} + \sqrt{5} = 3 + \sqrt{5}

Simplify

\sqrt{9}

to

3

to obtain the final simplified expression

Anahtar Kavram

Finding the square root of a surd expression by equating rational and radical parts

Soru 3838Soru →

A body is projected horizontally from the top of a cliff $45\text{ m}$ high. If it lands on flat ground at a horizontal distance of $120\text{ m}$ from the base of the cliff, what is the speed of the body just before it strikes the ground? (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 50

Cevap

The speed of the body just before striking the ground is 50 m/s.

The time of fall is determined by the height of

45\text{ m}

, yielding

t = \sqrt{2h/g} = 3\text{ s}

. The horizontal speed is constant at

120 / 3 = 40\text{ m/s}

. The vertical velocity gained on impact is

v_y = gt = 30\text{ m/s}

. Combining these mutually perpendicular velocity components gives a final impact speed of

v = \sqrt{40^2 + 30^2} = 50\text{ m/s}

.

Adım Adım Çözüm

1

Calculate the time of flight from the vertical height

t = 3 s

Vertical acceleration is constant under gravity while initial vertical velocity is zero.

2

Compute the constant horizontal component of velocity

v_x = 40 m/s

Horizontal speed is uniform because zero horizontal force acts on the projectile.

3

Compute the final vertical component of velocity at impact

v_y = 30 m/s

Vertical speed increases linearly with time due to gravitational acceleration.

4

Determine the magnitude of the resultant velocity vector

v = 50 m/s

The horizontal and vertical components are perpendicular, so their vector sum uses the Pythagorean theorem.

Anahtar Kavram

Horizontal Projection and Impact Velocity Vector

Tahmini Süre:1m 30s

Soru 3839Soru →

An iron ring has an internal cross-sectional area of $0.50\text{ m}^2$ at $30^\circ\text{C}$ . If the linear expansivity of iron is $1.2 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in its internal cross-sectional area when heated to $130^\circ\text{C}$ ?

1.2 \times 10^{-3}\text{ m}^2

6.0 \times 10^{-4}\text{ m}^2

1.8 \times 10^{-3}\text{ m}^2

2.4 \times 10^{-3}\text{ m}^2

Cevabı ve açıklamayı göster

Cevap: $1.2 \times 10^{-3}\text{ m}^2$

Cevap

The increase in the internal cross-sectional area of the ring is

1.2 \times 10^{-3}\text{ m}^2

.

For two-dimensional (area) expansion of solids, the area expansivity

\beta

is equal to twice the linear expansivity (

2\alpha

). Given

\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}

,

\beta = 2.4 \times 10^{-5}\text{ K}^{-1}

. Multiplying by the initial area (

0.50\text{ m}^2

) and temperature rise (

100\text{ K}

) yields an area increase of

1.2 \times 10^{-3}\text{ m}^2

.

Adım Adım Çözüm

1

Determine the temperature change (

\Delta T

) and the superficial expansivity (

\beta

).

\Delta T = 130^\circ\text{C} - 30^\circ\text{C} = 100\text{ K}

, and

\beta = 2\alpha = 2 \times (1.2 \times 10^{-5}\text{ K}^{-1}) = 2.4 \times 10^{-5}\text{ K}^{-1}

.

Area expansion depends on superficial expansivity, which is twice the linear expansivity for an isotropic solid.

2

Calculate the increase in area (

\Delta A

) using the area expansion formula.

\Delta A = A_0 \beta \Delta T = 0.50\text{ m}^2 \times (2.4 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 1.2 \times 10^{-3}\text{ m}^2

.

The fractional change in area is directly proportional to initial area, superficial expansivity, and temperature change.

Anahtar Kavram

Relationship between linear expansivity (

\alpha

) and superficial expansivity (

\beta = 2\alpha

) in thermal expansion of area.

Tahmini Süre:1m 30s

Soru 3840Soru →

A progressive sinusoidal wave traveling in a primary medium is governed by the equation $y = 0.08 \sin(120\pi t - 4\pi x)$ , where $x$ and $y$ are measured in meters and $t$ in seconds. When the wave passes into a second medium, its propagation speed changes to $15\text{ m/s}$ . What is the wavelength of the wave in the second medium?

0.50\text{ m}

0.25\text{ m}

0.125\text{ m}

4.00\text{ m}

Cevabı ve açıklamayı göster

Cevap: $0.25\text{ m}$

Cevap

0.25\text{ m}

Comparing the given equation

y = 0.08 \sin(120\pi t - 4\pi x)

to the standard wave equation

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

gives an angular frequency

\omega = 120\pi\text{ rad/s}

. The wave frequency is therefore

f = \frac{120\pi}{2\pi} = 60\text{ Hz}

. Since the frequency of a wave is determined solely by its source, it remains invariant when transmitting into a new medium. Given the new speed

v_2 = 15\text{ m/s}

, the new wavelength is

\lambda_2 = \frac{v_2}{f} = \frac{15}{60} = 0.25\text{ m}

.

Adım Adım Çözüm

1

Extract angular frequency (

\omega

) from the wave equation

Standard form

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

gives

\omega = 120\pi\text{ rad/s}

.

The coefficient of

t

in the argument of the sine function represents angular frequency.

2

Calculate the frequency of the wave in the primary medium

f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\text{ Hz}

.

Frequency

f

is related to angular frequency by

\omega = 2\pi f

.

3

Apply the principle of frequency invariance across boundary media

Frequency in the second medium

f_2 = 60\text{ Hz}

.

When a wave passes from one medium to another, its frequency depends only on the source and remains constant across boundaries.

4

Compute the wavelength in the second medium using the wave equation

v = f\lambda

\lambda_2 = \frac{v_2}{f_2} = \frac{15\text{ m/s}}{60\text{ Hz}} = 0.25\text{ m}

.

Wavelength is inversely proportional to frequency for a given wave speed in that medium.

Anahtar Kavram

Wave Equation Parameter Extraction and Frequency Invariance across Media Boundaries

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