Tüm alıştırma soruları

13931 soru

Soru 3841Soru

Which prepositions correctly complete the sentence below?

'The state governor declared that the proposed educational reforms were completely at variance _____ the aspirations of the local community and urged all stakeholders to desist _____ spreading misleading rumors.'

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Cevap: with / from

Cevap

The sentence is correctly completed by the pair 'with / from'.
The fixed prepositional phrase 'at variance with' expresses conflict or disagreement. Additionally, the verb 'desist' requires the preposition 'from' when specifying the action being stopped. Therefore, the option containing 'with / from' completes the sentence correctly.

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1
Determine the required preposition following the idiom 'at variance ____'.
The standard English idiom is 'at variance with', meaning in disagreement or opposition.
Fixed prepositional phrases require exact collocations.
2
Determine the dependent preposition governed by the verb 'desist ____'.
The verb 'desist' takes the complement preposition 'from'.
Certain verbs in formal English strictly govern specific prepositions.

Anahtar Kavram

Prepositional Idioms and Dependent Prepositions
Soru 3842Soru

A cell of electromotive force (e.m.f.) 1.5 V1.5\text{ V} and internal resistance 0.5 Ω0.5\text{ }\Omega is connected in series with a 2.5 Ω2.5\text{ }\Omega resistor. What is the terminal potential difference across the cell?

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Cevap: 1.25 V1.25\text{ V}

Cevap

The terminal potential difference across the cell is 1.25 V1.25\text{ V}.
When current flows through a closed circuit, voltage drops across both the external load resistor and the cell's internal resistance. Subtracting the lost voltage (0.25 V0.25\text{ V}) from the electromotive force (1.50 V1.50\text{ V}) yields a terminal potential difference of 1.25 V1.25\text{ V}.

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1
Calculate the total resistance of the circuit
Rtotal=R+r=2.5 Ω+0.5 Ω=3.0 ΩR_{\text{total}} = R + r = 2.5\text{ }\Omega + 0.5\text{ }\Omega = 3.0\text{ }\Omega
The external load resistor and the internal resistance of the cell are connected in series.
2
Calculate the total electric current flowing through the circuit
I=ER+r=1.5 V3.0 Ω=0.5 AI = \frac{E}{R + r} = \frac{1.5\text{ V}}{3.0\text{ }\Omega} = 0.5\text{ A}
Ohm's law for a complete circuit states that current equals total e.m.f. divided by total circuit resistance.
3
Determine the terminal potential difference across the cell
V=I×R=0.5 A×2.5 Ω=1.25 VV = I \times R = 0.5\text{ A} \times 2.5\text{ }\Omega = 1.25\text{ V}
Terminal potential difference is the potential drop across the external resistance.

Anahtar Kavram

Terminal Potential Difference and Cell Internal Resistance
Soru 3843Soru

An equilateral triangle has an area of 163 cm216\sqrt{3}\text{ cm}^2. If a square has the same perimeter as this equilateral triangle, what is the area of the square?

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Cevap: 36 cm236\text{ cm}^2

Cevap

The area of the square is 36 cm236\text{ cm}^2.
Using the area formula for an equilateral triangle Area=34s2=163\text{Area} = \frac{\sqrt{3}}{4}s^2 = 16\sqrt{3}, we determine the side length s=8 cms = 8\text{ cm}. The perimeter of the triangle is 3×8 cm=24 cm3 \times 8\text{ cm} = 24\text{ cm}. Since the square has an equal perimeter of 24 cm24\text{ cm}, each side of the square measures 24 cm/4=6 cm24\text{ cm} / 4 = 6\text{ cm}. The area of the square is therefore 62=36 cm26^2 = 36\text{ cm}^2.

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1
Find the side length of the equilateral triangle.
Side length s=8 cms = 8\text{ cm}.
The area of an equilateral triangle is given by Area=34s2\text{Area} = \frac{\sqrt{3}}{4}s^2. Setting 34s2=163\frac{\sqrt{3}}{4}s^2 = 16\sqrt{3} yields s2=64s^2 = 64, so s=8 cms = 8\text{ cm}.
2
Calculate the perimeter of the equilateral triangle.
Perimeter P=24 cmP = 24\text{ cm}.
An equilateral triangle has 3 equal sides, so P=3×8=24 cmP = 3 \times 8 = 24\text{ cm}.
3
Determine the side length of the square.
Square side length a=6 cma = 6\text{ cm}.
The square and triangle have equal perimeters (24 cm24\text{ cm}). Since a square has 4 equal sides, a=244=6 cma = \frac{24}{4} = 6\text{ cm}.
4
Calculate the area of the square.
Area =36 cm2= 36\text{ cm}^2.
The area of a square is a2=62=36 cm2a^2 = 6^2 = 36\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of Equilateral Triangles and Squares
Tahmini Süre:1m 30s
Soru 3844Soru

If 5+353535+3=x15\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = x\sqrt{15}, what is the value of xx?

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Cevap: 2

Cevap

The value of xx is 2.
Rationalising both fractions yields 4+154 + \sqrt{15} and 4154 - \sqrt{15}. Subtracting the second from the first gives (4+15)(415)=215(4 + \sqrt{15}) - (4 - \sqrt{15}) = 2\sqrt{15}. Comparing 2152\sqrt{15} with x15x\sqrt{15} gives x=2x = 2.

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1
Rationalise the denominator of the first fraction
\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = 4 + \sqrt{15}
Multiplying numerator and denominator by the conjugate of the denominator removes the surd from the denominator.
2
Rationalise the denominator of the second fraction
\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{5 - 2\sqrt{15} + 3}{5 - 3} = 4 - \sqrt{15}
Multiply by the conjugate (53)(\sqrt{5}-\sqrt{3}) to simplify the second surd term.
3
Subtract the simplified expressions
(4 + \sqrt{15}) - (4 - \sqrt{15}) = 4 - 4 + \sqrt{15} + \sqrt{15} = 2\sqrt{15}
Distribute the negative sign and combine like surd terms.
4
Solve for the unknown coefficient x
2\sqrt{15} = x\sqrt{15} \implies x = 2
Divide both sides of the equation by 15\sqrt{15} to isolate xx.

Anahtar Kavram

Binomial Surd Rationalisation and Simplification
Tahmini Süre:1m 30s
Soru 3845Soru

Complete the sentence below by supplying the correct prepositions to fill the blanks.

Aşağıdaki boşlukları doldurun

The investigative panel concluded that the managing director had acquiesced the fraudulent financial transactions and was entirely devoid remorse for his actions.
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Cevap

The correct prepositions are 'in' for the first blank and 'of' for the second blank.
The verb 'acquiesce' requires the dependent preposition 'in' to express tacit agreement or compliance, while the adjective 'devoid' consistently pairs with 'of' to mean completely lacking.

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1
Determine the correct dependent preposition that follows the verb 'acquiesce'.
The verb 'acquiesce' strictly collocates with the preposition 'in' when expressing passive compliance or agreement.
Using 'to' or 'with' after 'acquiesce' is a common prepositional error resulting from native language interference or erroneous analogy with 'agree to'.
2
Identify the standard preposition paired with the adjective 'devoid'.
The adjective 'devoid' takes the fixed preposition 'of' to denote being completely destitute or lacking in a quality.
Phrases such as 'devoid with' or 'devoid from' violate standard grammatical collocations in formal English.

Anahtar Kavram

Dependent Prepositions and Fixed Prepositional Collocations
Soru 3846Soru

Find the gradient of the normal to the curve y=x+1x1y = \frac{x + 1}{x - 1} at the point where x=3x = 3.

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Cevap: 2

Cevap

The gradient of the normal to the curve at x=3x = 3 is 22.
Differentiating y=x+1x1y = \frac{x + 1}{x - 1} via the quotient rule gives dydx=2(x1)2\frac{dy}{dx} = \frac{-2}{(x - 1)^2}. At x=3x = 3, the tangent gradient is mt=24=12m_t = \frac{-2}{4} = -\frac{1}{2}. Since the normal is perpendicular to the tangent, its gradient is mn=1mt=2m_n = -\frac{1}{m_t} = 2.

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1
Differentiate the rational function with respect to xx
Applying the quotient rule ddx(uv)=vuuvv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} where u=x+1u = x + 1 and v=x1v = x - 1 yields dydx=(x1)(1)(x+1)(1)(x1)2=2(x1)2\frac{dy}{dx} = \frac{(x - 1)(1) - (x + 1)(1)}{(x - 1)^2} = \frac{-2}{(x - 1)^2}.
The first derivative determines the slope function of the tangent line to the curve.
2
Substitute x=3x = 3 into the derivative to find the tangent slope mtm_t
mt=2(31)2=24=12m_t = \frac{-2}{(3 - 1)^2} = \frac{-2}{4} = -\frac{1}{2}.
Evaluating the derivative at the given xx-coordinate provides the exact gradient of the tangent at that point.
3
Calculate the slope of the normal line mnm_n
mn=1mt=11/2=2m_n = -\frac{1}{m_t} = -\frac{1}{-1/2} = 2.
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

Anahtar Kavram

The slope of the normal line to a curve at a given point is the negative reciprocal of the slope of the tangent line at that point (mn=1mtm_n = -\frac{1}{m_t}).
Soru 3847Soru

A sector of a circle of radius 21 cm21\text{ cm} has a total perimeter of 64 cm64\text{ cm}. Calculate the area of the sector in cm2\text{cm}^2. (Take π=227\pi = \frac{22}{7})

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Cevap: 231

Cevap

The area of the sector is 231 cm2231\text{ cm}^2.
The total perimeter of a sector is given by P=2r+lP = 2r + l. Given radius r=21 cmr = 21\text{ cm} and perimeter P=64 cmP = 64\text{ cm}, the arc length is l=642(21)=22 cml = 64 - 2(21) = 22\text{ cm}. The area of the sector is calculated using A=12rl=12×21×22=231 cm2A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2.

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1
Find the arc length of the sector
The arc length l=22 cml = 22\text{ cm}
The total perimeter of a sector includes its arc length plus its two bounding radii (P=2r+lP = 2r + l). Subtracting 2r=42 cm2r = 42\text{ cm} from 64 cm64\text{ cm} gives l=22 cml = 22\text{ cm}.
2
Calculate the area of the sector
The area A=231 cm2A = 231\text{ cm}^2
Using the relation between arc length and sector area, A=12rl=12×21×22=231 cm2A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of a Sector of a Circle
Tahmini Süre:1m 30s
Soru 3848Soru

The committee chairman noted that although several board members were initially skeptical of the restructuring policy, they eventually proved amenable _____ the proposed changes. Which preposition correctly completes the sentence above?

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Cevap: to

Cevap

The correct option is 'to' because the adjective 'amenable' requires the preposition 'to' when signifying open, responsive, or willing to yield.
In standard formal English, the adjective 'amenable' takes the dependent preposition 'to' when used to mean open, receptive, or agreeable to a plan or advice. Therefore, 'to' is the only grammatically correct choice.

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1
Identify the head adjective requiring a dependent preposition in the sentence context.
The word requiring a complement preposition is the predicate adjective 'amenable'.
Certain adjectives in English take obligatory fixed dependent prepositions to complete their meaning.
2
Recall standard prepositional collocation rules for 'amenable'.
The correct complement structure is 'amenable to something' (meaning responsive, willing, or agreeable to something).
Substituting other prepositions such as 'with', 'for', or 'about' violates standard grammatical usage in formal English.

Anahtar Kavram

Dependent Prepositions and Adjective-Preposition Collocations
Soru 3849Soru

If x+4x1=1\sqrt{x + 4} - \sqrt{x - 1} = 1, what is the value of xx?

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Cevap: 55

Cevap

The value of xx is 55.
Isolating x+4\sqrt{x + 4} gives x+4=1+x1\sqrt{x + 4} = 1 + \sqrt{x - 1}. Squaring both sides yields x+4=1+2x1+x1x + 4 = 1 + 2\sqrt{x - 1} + x - 1, which simplifies to 4=2x14 = 2\sqrt{x - 1}. Dividing by 2 gives 2=x12 = \sqrt{x - 1}. Squaring both sides once more gives 4=x14 = x - 1, which leads to x=5x = 5.

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1
Isolate one of the surd terms on the left side of the equation.
x+4=1+x1\sqrt{x + 4} = 1 + \sqrt{x - 1}
Rearranging terms prevents dealing with complex cross-products when squaring.
2
Square both sides of the equation.
x+4=1+2x1+(x1)x + 4 = 1 + 2\sqrt{x - 1} + (x - 1)
Squaring eliminates the outer radical on the left side.
3
Simplify both sides and isolate the remaining radical term.
4=2x1    2=x14 = 2\sqrt{x - 1} \implies 2 = \sqrt{x - 1}
Subtracting xx from both sides simplifies the algebraic expression.
4
Square both sides again to solve for xx.
4=x1    x=54 = x - 1 \implies x = 5
Squaring eliminates the remaining radical term.

Anahtar Kavram

Solving Surd Equations by Rational Elimination
Soru 3850Soru

Monochromatic light of wavelength 500 nm500\text{ nm} is incident normally on a plane diffraction grating. If the second-order principal maximum is observed at an angle of 3030^\circ to the normal, calculate the number of lines per millimeter ruled on the grating.

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Cevap: 500

Cevap

500 lines/mm
Using the diffraction grating equation dsinθ=nλd \sin\theta = n\lambda with n=2n = 2, λ=5.0×107 m\lambda = 5.0 \times 10^{-7}\text{ m}, and sin(30)=0.5\sin(30^\circ) = 0.5 gives a slit separation of d=2.0×106 md = 2.0 \times 10^{-6}\text{ m}. Converting to grating ruling density per millimeter yields 103 m2.0×106 m=500 lines/mm\frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}.

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1
Identify the given physical parameters and convert wavelength to standard SI meters.
Order n=2n = 2, diffraction angle θ=30\theta = 30^\circ, wavelength λ=500×109 m=5.0×107 m\lambda = 500 \times 10^{-9}\text{ m} = 5.0 \times 10^{-7}\text{ m}.
Standard SI unit conversion is required to perform wave equation calculations accurately.
2
Apply the diffraction grating equation dsinθ=nλd \sin\theta = n\lambda to compute the grating spacing dd.
d=2×5.0×107 msin30=1.0×1060.5=2.0×106 md = \frac{2 \times 5.0 \times 10^{-7}\text{ m}}{\sin 30^\circ} = \frac{1.0 \times 10^{-6}}{0.5} = 2.0 \times 10^{-6}\text{ m}.
The condition for principal constructive interference maxima is dsinθ=nλd \sin\theta = n\lambda.
3
Calculate the number of lines per millimeter by dividing 1 mm1\text{ mm} (103 m10^{-3}\text{ m}) by the grating spacing dd.
Nmm=103 m2.0×106 m=500 lines/mmN_{\text{mm}} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}.
Grating density (lines per unit length) is the reciprocal of the slit separation dd.

Anahtar Kavram

Diffraction Grating Principal Maxima Condition
Tahmini Süre:2m 0s
Soru 3851Soru

Given the universal set U={xZ:1x15}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}, with subsets P={xU:x is a prime number}P = \{x \in \mathcal{U} : x \text{ is a prime number}\} and Q={xU:x is an odd number}Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}, what is the number of elements in the set (PQ)(P \cup Q)'?

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Cevap: 6; six; 6 elements

Cevap

The number of elements in (PQ)(P \cup Q)' is 6.
The set PQP \cup Q consists of all numbers from 1 to 15 that are either prime or odd: {1,2,3,5,7,9,11,13,15}\{1, 2, 3, 5, 7, 9, 11, 13, 15\}. The complement (PQ)(P \cup Q)' relative to U\mathcal{U} contains all elements of U\mathcal{U} that are neither prime nor odd, which are the even composite numbers: {4,6,8,10,12,14}\{4, 6, 8, 10, 12, 14\}. Counting these elements gives 6.

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1
List all elements of the universal set U\mathcal{U}, subset PP, and subset QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}, P={2,3,5,7,11,13}P = \{2, 3, 5, 7, 11, 13\}, and Q={1,3,5,7,9,11,13,15}Q = \{1, 3, 5, 7, 9, 11, 13, 15\}.
Listing the explicit elements allows precise execution of set union and complement operations.
2
Find the union PQP \cup Q.
PQ={1,2,3,5,7,9,11,13,15}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}.
The union combines all unique elements that belong to either set PP, set QQ, or both.
3
Determine the complement set (PQ)(P \cup Q)' relative to U\mathcal{U} and count its elements.
(PQ)={4,6,8,10,12,14}(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}, which contains 6 elements.
The complement set consists of all elements in U\mathcal{U} that are not present in PQP \cup Q.

Anahtar Kavram

Complement of Set Union
Soru 3852Soru

An electric heater of resistance 20Ω20\,\Omega carries a current of 5A5\,\text{A} when operated. What is the electrical power dissipated by the heater?

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Cevap: 500W500\,\text{W}

Cevap

500W500\,\text{W}
According to Joule's law of electric heating, electrical power PP dissipated in a resistor of resistance RR carrying current II is given by P=I2RP = I^2 R. Substituting the given values I=5AI = 5\,\text{A} and R=20ΩR = 20\,\Omega gives P=(5A)2×20Ω=25×20=500WP = (5\,\text{A})^2 \times 20\,\Omega = 25 \times 20 = 500\,\text{W}.

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1
Identify the given quantities from the problem statement
Current I=5AI = 5\,\text{A} and resistance R=20ΩR = 20\,\Omega
These parameters are required to calculate the power dissipated.
2
Select the appropriate formula for power in terms of current and resistance
P=I2RP = I^2 R
Joule's heating law specifies that power dissipation is given by P=I2RP = I^2 R.
3
Substitute the values and evaluate the numerical result
P=(5)2×20=25×20=500WP = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}
Squaring 55 yields 2525, and multiplying by 2020 gives the final power output of 500W500\,\text{W}.

Anahtar Kavram

Electrical Power Dissipation in a Resistor
Tahmini Süre:45s
Soru 3853Soru

An electric current of 2.0A2.0\,\text{A} flows through a resistor of resistance 5.0Ω5.0\,\Omega for a duration of 10.0seconds10.0\,\text{seconds}. What is the total electrical energy, in Joules, dissipated by the resistor?

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Cevap: 200

Cevap

The total electrical energy dissipated by the resistor is 200J200\,\text{J}.
According to Joule's law of heating, the electrical energy EE converted into thermal energy when a current II flows through a resistor RR for time tt is given by E=I2RtE = I^2 R t. Substituting I=2.0AI = 2.0\,\text{A}, R=5.0ΩR = 5.0\,\Omega, and t=10.0st = 10.0\,\text{s} gives E=(2.0)2×5.0×10.0=4.0×5.0×10.0=200JE = (2.0)^2 \times 5.0 \times 10.0 = 4.0 \times 5.0 \times 10.0 = 200\,\text{J}.

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1
Identify the given physical values from the problem statement.
I=2.0AI = 2.0\,\text{A}, R=5.0ΩR = 5.0\,\Omega, t=10.0st = 10.0\,\text{s}.
These are the necessary parameters to compute electrical energy.
2
State the formula for electrical energy dissipated in a resistor (Joule's law).
E=I2RtE = I^2 R t
Electrical energy is equal to power multiplied by time, where power P=I2RP = I^2 R.
3
Calculate the numerical value of the energy.
E=(2.0)2×5.0×10.0=200JE = (2.0)^2 \times 5.0 \times 10.0 = 200\,\text{J}
Squaring the current gives 4.0A24.0\,\text{A}^2, and multiplying by 5.0Ω5.0\,\Omega and 10.0s10.0\,\text{s} yields 200J200\,\text{J}.

Anahtar Kavram

Joule's Law of Electrical Heating
Soru 3854Soru

A 6 μF6\text{ }\mu\text{F} capacitor is connected in series with a parallel arrangement of a 2 μF2\text{ }\mu\text{F} capacitor and a 1 μF1\text{ }\mu\text{F} capacitor. If the entire circuit is connected across a 30 V30\text{ V} d.c. power supply, what is the electric charge stored on the 2 μF2\text{ }\mu\text{F} capacitor?

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Cevap: 40 μC40\text{ }\mu\text{C}

Cevap

The charge stored on the 2 μF2\text{ }\mu\text{F} capacitor is 40 μC40\text{ }\mu\text{C}.
To find the charge on the 2 μF2\text{ }\mu\text{F} capacitor, we first combine the parallel capacitors (2 μF+1 μF=3 μF2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}). Next, we combine this in series with the 6 μF6\text{ }\mu\text{F} capacitor to get an overall capacitance Ceq=2 μFC_{eq} = 2\text{ }\mu\text{F}. The total charge drawn from the 30 V30\text{ V} supply is Q=2 μF×30 V=60 μCQ = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}. This total charge enters the parallel combination, giving a potential drop across the parallel branch of Vp=60 μC/3 μF=20 VV_p = 60\text{ }\mu\text{C} / 3\text{ }\mu\text{F} = 20\text{ V}. Therefore, the charge on the 2 μF2\text{ }\mu\text{F} capacitor is 2 μF×20 V=40 μC2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}.

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1
Calculate the equivalent capacitance of the parallel section
Cp=2 μF+1 μF=3 μFC_{p} = 2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}
Capacitors in parallel add directly.
2
Calculate the total equivalent capacitance of the network
Ceq=6×36+3=189=2 μFC_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\text{ }\mu\text{F}
The 6 μF6\text{ }\mu\text{F} capacitor and the 3 μF3\text{ }\mu\text{F} parallel equivalent are in series.
3
Find the total charge supplied by the 30 V30\text{ V} source
Qtotal=Ceq×V=2 μF×30 V=60 μCQ_{total} = C_{eq} \times V = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}
Total charge is the product of equivalent capacitance and total voltage.
4
Determine the potential difference across the parallel branch
Vp=QtotalCp=60 μC3 μF=20 VV_{p} = \frac{Q_{total}}{C_{p}} = \frac{60\text{ }\mu\text{C}}{3\text{ }\mu\text{F}} = 20\text{ V}
The total charge flows through the series combination, creating a potential drop across the parallel combination equal to Qtotal/CpQ_{total} / C_{p}.
5
Calculate the charge on the 2 μF2\text{ }\mu\text{F} capacitor
Q2μF=C2μF×Vp=2 μF×20 V=40 μCQ_{2\mu\text{F}} = C_{2\mu\text{F}} \times V_{p} = 2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}
The charge on a specific capacitor in parallel is the product of its capacitance and the voltage across the parallel branch.

Anahtar Kavram

Charge and voltage distribution in mixed capacitor networks
Soru 3855Soru

A sports field consists of a central rectangular section of length 100 m100\text{ m} bounded on two opposite ends by semicircular regions, each having a radius of 35 m35\text{ m}. What is the total area of the sports field? (Take π=227\pi = \frac{22}{7})

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Cevap: 10,850 m210,850\text{ m}^2

Cevap

The total area of the sports field is 10,850 m210,850\text{ m}^2.
The sports field is composed of a central rectangle of dimensions 100 m100\text{ m} by 70 m70\text{ m} (since width is equal to the diameter 2×35 m=70 m2 \times 35\text{ m} = 70\text{ m}) and two semicircular ends of radius 35 m35\text{ m}. The rectangular area is 100×70=7,000 m2100 \times 70 = 7,000\text{ m}^2. The two semicircles join to make one full circle with an area of 227×352=3,850 m2\frac{22}{7} \times 35^2 = 3,850\text{ m}^2. Adding both parts gives 7,000+3,850=10,850 m27,000 + 3,850 = 10,850\text{ m}^2.

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1
Calculate the width of the rectangular region.
Width = 2×r=2×35 m=70 m2 \times r = 2 \times 35\text{ m} = 70\text{ m}.
The diameter of the semicircular ends forms the width of the central rectangle.
2
Calculate the area of the rectangular region.
Area of rectangle = length×width=100 m×70 m=7,000 m2\text{length} \times \text{width} = 100\text{ m} \times 70\text{ m} = 7,000\text{ m}^2.
Formula for the area of a rectangle is length×width\text{length} \times \text{width}.
3
Calculate the combined area of the two semicircular ends.
Combined area = πr2=227×35×35=22×5×35=3,850 m2\pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3,850\text{ m}^2.
Two identical semicircles of radius rr combine to form one full circle of radius rr.
4
Add the rectangular area and the combined circular area.
Total Area = 7,000 m2+3,850 m2=10,850 m27,000\text{ m}^2 + 3,850\text{ m}^2 = 10,850\text{ m}^2.
The total area is the sum of the composite plane shapes.

Anahtar Kavram

Area of Composite Plane Figures
Soru 3856Soru

A curve is defined by the equation y=x33x2+ky = x^3 - 3x^2 + k, where kk is a constant. If the local minimum value of yy on the curve is 22, what is the value of kk?

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Cevap: 6

Cevap

The value of the constant kk is 66.
To find the constant kk, differentiate the curve equation to obtain dydx=3x26x\frac{dy}{dx} = 3x^2 - 6x. Setting dydx=0\frac{dy}{dx} = 0 gives stationary points at x=0x = 0 and x=2x = 2. Calculating the second derivative d2ydx2=6x6\frac{d^2y}{dx^2} = 6x - 6 shows d2ydx2=6>0\frac{d^2y}{dx^2} = 6 > 0 at x=2x = 2, confirming that the local minimum occurs at x=2x = 2. Substituting x=2x = 2 and the minimum value y=2y = 2 into y=x33x2+ky = x^3 - 3x^2 + k yields 2=812+k2 = 8 - 12 + k, which simplifies to k=6k = 6.

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1
Differentiate y=x33x2+ky = x^3 - 3x^2 + k with respect to xx.
dydx=3x26x\frac{dy}{dx} = 3x^2 - 6x
Stationary points occur where the first derivative equals zero.
2
Solve dydx=0\frac{dy}{dx} = 0 for xx.
x=0x = 0 or x=2x = 2
These xx-values locate the turning points on the curve.
3
Evaluate the second derivative d2ydx2=6x6\frac{d^2y}{dx^2} = 6x - 6 at each stationary point.
At x=0x = 0, d2ydx2=6\frac{d^2y}{dx^2} = -6 (local maximum); at x=2x = 2, d2ydx2=6\frac{d^2y}{dx^2} = 6 (local minimum).
A positive second derivative indicates a local minimum point.
4
Substitute x=2x = 2 and y=2y = 2 into the curve equation y=x33x2+ky = x^3 - 3x^2 + k.
2=(2)33(2)2+k    2=4+k    k=62 = (2)^3 - 3(2)^2 + k \implies 2 = -4 + k \implies k = 6
The local minimum value of yy is attained at x=2x = 2.

Anahtar Kavram

Stationary Points, Maxima, and Minima
Soru 3857Soru

Which prepositions correctly complete the sentence below?

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The university senate insisted that all postgraduate applications must conform the newly approved admission guidelines, irrespective the applicants' prior qualifications.
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Cevap

The correct prepositions needed to complete the sentence are 'to' for the first blank and 'of' for the second blank.
The verb 'conform' requires the dependent preposition 'to' when indicating adherence to established standards or guidelines. Additionally, 'irrespective' forms a fixed prepositional phrase with 'of' ('irrespective of') to mean 'without taking into account'. Thus, 'to' and 'of' correctly complete the sentence.

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1
Determine the correct dependent preposition required after the verb 'conform' when referring to rules or standards.
The verb 'conform' collocates with the preposition 'to'.
In standard English usage, one conforms 'to' a set of rules, regulations, or guidelines.
2
Identify the standard multi-word prepositional phrase starting with 'irrespective'.
The fixed prepositional phrase is 'irrespective of'.
'Irrespective' requires the preposition 'of' to form a functioning compound preposition meaning 'regardless of'.

Anahtar Kavram

Prepositions and Prepositional Phrases
Tahmini Süre:1m 0s
Soru 3858Soru

An athlete throws a javelin from ground level such that its initial vertical component of velocity is 40 m/s40\text{ m/s} and its initial horizontal component of velocity is 30 m/s30\text{ m/s}. What is the horizontal distance in metres covered by the javelin when it reaches a height of 35 m35\text{ m} above the ground for the first time? (Take g=10 m/s2g = 10\text{ m/s}^2)

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Cevap: 30

Cevap

The horizontal distance covered by the javelin when it reaches a height of 35 m for the first time is 30 m.
Using y=uyt12gt2y = u_y t - \frac{1}{2}g t^2 with uy=40 m/su_y = 40\text{ m/s}, y=35 my = 35\text{ m}, and g=10 m/s2g = 10\text{ m/s}^2 yields 35=40t5t235 = 40t - 5t^2. Dividing by 5 gives t28t+7=0t^2 - 8t + 7 = 0, which factors to (t1)(t7)=0(t - 1)(t - 7) = 0. The roots are t=1 st = 1\text{ s} (ascent) and t=7 st = 7\text{ s} (descent). For the first time, t=1 st = 1\text{ s}. The horizontal displacement is x=uxt=30 m/s×1 s=30 mx = u_x t = 30\text{ m/s} \times 1\text{ s} = 30\text{ m}.

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1
Set up the vertical motion equation to find the time when height is 35 m
35=40t5t235 = 40t - 5t^2
Vertical displacement in projectile motion depends on the vertical initial velocity component and acceleration due to gravity.
2
Solve the quadratic equation for time tt
t28t+7=0    t=1 s or t=7 st^2 - 8t + 7 = 0 \implies t = 1\text{ s} \text{ or } t = 7\text{ s}
A projectile reaches a given non-peak height twice: once ascending and once descending.
3
Select the first time value and calculate horizontal distance
x=ux×t=30×1=30 mx = u_x \times t = 30 \times 1 = 30\text{ m}
Horizontal velocity remains constant throughout the flight, so distance is speed multiplied by time.

Anahtar Kavram

Independence of vertical and horizontal components in projectile motion
Soru 3859Soru

A uniform metal wire of length 20m20\,\text{m} and total mass 0.034kg0.034\,\text{kg} is manufactured from a material of density 8.5×103kg/m38.5 \times 10^3\,\text{kg/m}^3 and electrical resistivity 1.7×108Ωm1.7 \times 10^{-8}\,\Omega\cdot\text{m}. What is the electrical resistance of the wire in ohms?

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Cevap: 1.7

Cevap

The electrical resistance of the wire is 1.7Ω1.7\,\Omega.
Combining the density relation V=mdV = \frac{m}{d} with the geometric expression V=ALV = A \cdot L gives A=mdLA = \frac{m}{d L}. Substituting this into Pouillet's law R=ρLAR = \frac{\rho L}{A} yields R=ρdL2mR = \frac{\rho d L^2}{m}. Evaluating with the given values: R=(1.7×108)(8.5×103)(20)20.034=1.7ΩR = \frac{(1.7 \times 10^{-8})(8.5 \times 10^3)(20)^2}{0.034} = 1.7\,\Omega.

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1
Calculate the volume of the wire using mass and density
V=4.0×106m3V = 4.0 \times 10^{-6}\,\text{m}^3
Volume is related to mass and density by V=mdV = \frac{m}{d}.
2
Calculate the cross-sectional area of the wire
A=2.0×107m2A = 2.0 \times 10^{-7}\,\text{m}^2
For a cylindrical wire of uniform cross-section, V=ALV = A \cdot L, so A=VLA = \frac{V}{L}.
3
Apply resistivity formula to find electrical resistance
R=1.7ΩR = 1.7\,\Omega
Resistance is given by R=ρLAR = \frac{\rho L}{A}.

Anahtar Kavram

Relationship between Resistance, Mass, Density, and Resistivity
Tahmini Süre:2m 0s
Soru 3860Soru

An aerosol canister containing an ideal gas at an initial pressure of 2.20×105 Pa2.20 \times 10^5\text{ Pa} and a temperature of 27C27^\circ\text{C} is accidentally thrown into a fire, causing the temperature of the gas to rise to 327C327^\circ\text{C}. Assuming the volume of the canister remains constant, what is the new pressure of the gas inside the canister?

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Cevap: 4.40×105 Pa4.40 \times 10^5\text{ Pa}

Cevap

The new pressure of the gas inside the canister is 4.40×105 Pa4.40 \times 10^5\text{ Pa}.
According to Pressure's Law, the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature in Kelvin (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}). Converting the temperatures gives T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=327+273=600 KT_2 = 327 + 273 = 600\text{ K}. Multiplying the initial pressure by the ratio 600 K300 K=2\frac{600\text{ K}}{300\text{ K}} = 2 gives 4.40×105 Pa4.40 \times 10^5\text{ Pa}.

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1
Convert initial and final temperatures from degrees Celsius to Kelvin.
T1=27C+273=300 KT_1 = 27^\circ\text{C} + 273 = 300\text{ K} and T2=327C+273=600 KT_2 = 327^\circ\text{C} + 273 = 600\text{ K}.
All calculations using gas laws require absolute temperature measured in Kelvin.
2
State and rearrange Pressure's Law (Gay-Lussac's Law) for constant volume.
P1T1=P2T2    P2=P1×T2T1\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}.
For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.
3
Substitute the known values to calculate the final pressure P2P_2.
P2=2.20×105 Pa×600 K300 K=4.40×105 PaP_2 = 2.20 \times 10^5\text{ Pa} \times \frac{600\text{ K}}{300\text{ K}} = 4.40 \times 10^5\text{ Pa}.
Since absolute temperature doubles from 300 K300\text{ K} to 600 K600\text{ K}, the final pressure must also double.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law) and Absolute Temperature Conversion
Tahmini Süre:1m 30s
ÖncekiSayfa 193 / 697Sonraki
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