Tüm alıştırma soruları

13931 soru

Soru 4041Soru

An AC series circuit consists of a resistor of resistance R=6 ΩR = 6\ \Omega connected in series with an inductor of inductive reactance XL=8 ΩX_L = 8\ \Omega. What is the total impedance of the circuit?

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Cevap: 10 Ω10\ \Omega

Cevap

The total impedance of the circuit is 10 Ω10\ \Omega.
The net impedance ZZ in an AC series circuit containing a resistor and an inductor is given by Z=R2+XL2Z = \sqrt{R^2 + X_L^2}. Substituting R=6 ΩR = 6\ \Omega and XL=8 ΩX_L = 8\ \Omega yields Z=62+82=36+64=100=10 ΩZ = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \Omega.

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1
Identify the formula for impedance in an RL series circuit
Z=R2+XL2Z = \sqrt{R^2 + X_L^2}
In an AC circuit, resistance and inductive reactance add vectorially at a 9090^\circ phase angle.
2
Substitute the given values R=6 ΩR = 6\ \Omega and XL=8 ΩX_L = 8\ \Omega into the formula
Z=62+82=36+64=100Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}
Squaring each component gives the terms required under the radical.
3
Calculate the square root
Z=10 ΩZ = 10\ \Omega
The square root of 100 gives the net opposition to current flow (impedance).

Anahtar Kavram

Impedance of an RL Series AC Circuit
Tahmini Süre:45s
Soru 4042Soru

What quantity of heat energy is required to completely melt 0.50 kg0.50\text{ kg} of ice at 0C0^\circ\text{C} into water at 0C0^\circ\text{C}? (Specific latent heat of fusion of ice = 3.3×105 J kg13.3 \times 10^5\text{ J kg}^{-1})

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Cevap: 1.65×105 J1.65 \times 10^5\text{ J}

Cevap

The quantity of heat energy required is 1.65×105 J1.65 \times 10^5\text{ J}.
The quantity of heat required for melting at constant temperature depends only on mass and specific latent heat of fusion: Q=mLf=0.50×3.3×105=1.65×105 JQ = mL_f = 0.50 \times 3.3 \times 10^5 = 1.65 \times 10^5\text{ J}.

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1
Identify the given values and formula for change of state
Mass m=0.50 kgm = 0.50\text{ kg}, Specific latent heat of fusion Lf=3.3×105 J kg1L_f = 3.3 \times 10^5\text{ J kg}^{-1}. The formula is Q=mLfQ = m L_f.
During a phase change at constant temperature, latent heat is absorbed without a change in temperature.
2
Calculate total heat energy QQ
Q=0.50 kg×3.3×105 J kg1=1.65×105 JQ = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J kg}^{-1} = 1.65 \times 10^5\text{ J}.
Multiplying mass by the specific latent heat gives the total energy transferred.

Anahtar Kavram

Latent Heat of Fusion
Tahmini Süre:45s
Soru 4043Soru

Find the sum, in degrees, of all solutions to the trigonometric equation 3tan(2x)=3\sqrt{3}\tan(2x) = 3 in the interval 0x1800^\circ \le x \le 180^\circ.

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Cevap: 150

Cevap

The sum of all solutions to the equation in the given interval is 150 degrees.
Isolating tan(2x)\tan(2x) gives 3\sqrt{3}. For 02x3600^\circ \le 2x \le 360^\circ, tan(2x)=3\tan(2x) = \sqrt{3} yields solutions at 2x=602x = 60^\circ and 2x=2402x = 240^\circ. Dividing by 2 gives x=30x = 30^\circ and x=120x = 120^\circ. Adding these solutions yields 30+120=15030^\circ + 120^\circ = 150^\circ.

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1
Isolate the trigonometric function
tan(2x)=33=3\tan(2x) = \frac{3}{\sqrt{3}} = \sqrt{3}
Dividing both sides by \sqrt{3} simplifies the expression to a standard special angle ratio.
2
Determine the domain for the argument 2x2x
Since 0x1800^\circ \le x \le 180^\circ, multiplying the inequality by 2 gives 02x3600^\circ \le 2x \le 360^\circ.
This establishes the range of angles to search for 2x2x within one complete turn.
3
Find all values of 2x2x where tangent equals 3\sqrt{3}
2x=602x = 60^\circ (1st quadrant) and 2x=180+60=2402x = 180^\circ + 60^\circ = 240^\circ (3rd quadrant)
The tangent function is positive in Quadrants I and III with a reference angle of 6060^\circ.
4
Solve for xx
x=602=30x = \frac{60^\circ}{2} = 30^\circ and x=2402=120x = \frac{240^\circ}{2} = 120^\circ
Dividing each angle by 2 yields the values of xx lying within the domain 0x1800^\circ \le x \le 180^\circ.
5
Calculate the sum of the solutions
30+120=15030^\circ + 120^\circ = 150^\circ
The question specifically requests the sum of all valid solutions.

Anahtar Kavram

Solving trigonometric equations using reference angles and domain transformation
Tahmini Süre:2m 0s
Soru 4044Soru

In ΔABC\Delta ABC, side a=6 cma = 6\text{ cm}, side b=10 cmb = 10\text{ cm}, and the included angle C=120\angle C = 120^\circ. What is the length of side cc?

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Cevap: 14 cm14\text{ cm}

Cevap

The length of side cc is 14 cm14\text{ cm}.
According to the Cosine Rule c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C, substituting the given values a=6a = 6, b=10b = 10, and cos120=12\cos 120^\circ = -\frac{1}{2} yields c2=36+1002(6)(10)(12)=136+60=196c^2 = 36 + 100 - 2(6)(10)\left(-\frac{1}{2}\right) = 136 + 60 = 196. Taking the positive square root gives c=14 cmc = 14\text{ cm}.

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1
Identify the given values and state the relevant Cosine Rule formula
Given: a=6 cma = 6\text{ cm}, b=10 cmb = 10\text{ cm}, C=120\angle C = 120^\circ. Formula: c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C.
Since two sides and the included angle (SAS configuration) are known, the Cosine Rule must be used to find the third side.
2
Evaluate cos120\cos 120^\circ and substitute all values into the formula
cos120=12\cos 120^\circ = -\frac{1}{2}. Thus, c2=62+1022(6)(10)(12)c^2 = 6^2 + 10^2 - 2(6)(10)\left(-\frac{1}{2}\right).
Cosine of an obtuse angle in the second quadrant is negative.
3
Simplify the algebraic expression
c2=36+100+60=196c^2 = 36 + 100 + 60 = 196.
Multiplying 2(60)(12)-2(60)\left(-\frac{1}{2}\right) yields +60+60.
4
Take the principal square root to solve for cc
c=196=14 cmc = \sqrt{196} = 14\text{ cm}.
Length must be a positive real number.

Anahtar Kavram

Cosine Rule for finding an unknown side in SAS triangle configurations
Soru 4045Soru

In how many different ways can the letters of the word SUCCESS be arranged such that the three 'S's do not all come together?

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Cevap: 360

Cevap

The letters of the word SUCCESS can be arranged in 360 ways such that the three 'S's do not all come together.
The correct answer is calculated using complementary counting. First, the total unrestricted permutations of SUCCESS (7 letters with 3 'S's and 2 'C's) is 7!3!2!=420\frac{7!}{3!2!} = 420. Next, treating the three 'S's as one single block leaves 5 items to arrange with 2 'C's, giving 5!2!=60\frac{5!}{2!} = 60 ways where the 'S's are together. Subtracting 60 from 420 yields 360.

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1
Calculate the total number of unrestricted arrangements of the word SUCCESS.
The word SUCCESS has 7 letters in total: 3 'S's, 2 'C's, 1 'U', and 1 'E'. Total arrangements Ntotal=7!3!×2!=50406×2=420N_{total} = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420.
Repeated letters must be accounted for by dividing the factorial of the total count by the factorials of the counts of repeated letters.
2
Calculate the number of arrangements where the three 'S's are all together.
Treat the three 'S's as a single entity (SSS). We now arrange 5 entities: (SSS), U, C, C, E. Since 'C' appears twice, Ntogether=5!2!=1202=60N_{together} = \frac{5!}{2!} = \frac{120}{2} = 60.
Grouping restricted identical items into a single block allows us to find the subset of arrangements where they stay together.
3
Subtract the number of 'together' arrangements from the total arrangements.
Nnot_together=NtotalNtogether=42060=360N_{not\_together} = N_{total} - N_{together} = 420 - 60 = 360.
The complementary counting principle gives the number of ways where the restriction is satisfied.

Anahtar Kavram

Permutations with Repeated Elements and Complementary Counting
Tahmini Süre:1m 30s
Soru 4046Soru

A metal rod of length 12.5 cm12.5\text{ cm} is divided into two parts in the ratio 2:32 : 3. If the length of the smaller part is measured as 5.2 cm5.2\text{ cm}, what is the percentage error in the measurement of the smaller part?

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Cevap: 4.0%4.0\%

Cevap

The percentage error in the measurement of the smaller part is 4.0%4.0\%.
The true length of the smaller part of the rod is 25×12.5=5.0 cm\frac{2}{5} \times 12.5 = 5.0\text{ cm}. Subtracting this from the measured length 5.2 cm5.2\text{ cm} gives an error of 0.2 cm0.2\text{ cm}. Dividing the error 0.2 cm0.2\text{ cm} by the true value 5.0 cm5.0\text{ cm} and multiplying by 100%100\% yields 4.0%4.0\%.

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1
Calculate the actual length of the smaller part using the given ratio.
Total ratio parts = 2+3=52 + 3 = 5. Actual length of smaller part = 25×12.5 cm=5.0 cm\frac{2}{5} \times 12.5\text{ cm} = 5.0\text{ cm}.
The total length is shared according to the ratio 2:32 : 3, so the smaller piece corresponds to 2 out of 5 equal parts.
2
Find the absolute error in measurement.
Absolute error = 5.2 cm5.0 cm=0.2 cm|5.2\text{ cm} - 5.0\text{ cm}| = 0.2\text{ cm}.
Absolute error is the magnitude of the difference between the measured value and the true value.
3
Compute the percentage error.
Percentage error = Absolute ErrorTrue Value×100%=0.25.0×100%=4.0%\frac{\text{Absolute Error}}{\text{True Value}} \times 100\% = \frac{0.2}{5.0} \times 100\% = 4.0\%.
Percentage error is defined as the absolute error divided by the true value, multiplied by 100%.

Anahtar Kavram

Ratio partitioning and Percentage Error calculation
Tahmini Süre:1m 30s
Soru 4047Soru

A trader invested 50,000\text{₦}50,000 in a savings scheme that pays compound interest at a rate of 10%10\% per annum compounded annually. What is the total compound interest earned at the end of 22 years?

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Cevap: 10,500\text{₦}10,500

Cevap

The total compound interest earned at the end of 22 years is 10,500\text{₦}10,500.
The compound interest is obtained by subtracting the principal from the total accumulated amount. Using A=P(1+r)nA = P(1 + r)^n, the total amount is 60,500\text{₦}60,500. Subtracting the principal of 50,000\text{₦}50,000 yields 10,500\text{₦}10,500.

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1
Calculate the total accumulated amount using the compound interest formula A=P(1+r)nA = P(1 + r)^n
A=50,000×(1+0.10)2=50,000×1.21=60,500A = 50,000 \times (1 + 0.10)^2 = 50,000 \times 1.21 = \text{₦}60,500
Determines the total value of the investment at the end of the duration.
2
Calculate the compound interest earned using CI=APCI = A - P
CI=60,50050,000=10,500CI = 60,500 - 50,000 = \text{₦}10,500
Subtracts the original principal from the total accumulated amount to find the interest portion.

Anahtar Kavram

Compound interest calculation
Soru 4048Soru

Given the universal set U={xZ:1x20}U = \{x \in \mathbb{Z} : 1 \le x \le 20\}. Let P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\}, Q={xU:x is an odd integer}Q = \{x \in U : x \text{ is an odd integer}\}, and R={xU:x is a multiple of 3}R = \{x \in U : x \text{ is a multiple of } 3\}. What is the value of n((PQ)R)n((P \cup Q) \cap R')?

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Cevap: 8

Cevap

8
The correct value is 8. The union PQP \cup Q yields {1,2,3,5,7,9,11,13,15,17,19}\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}. Intersecting this set with RR' means removing any element that is a multiple of 3. The multiples of 3 in PQP \cup Q are 3, 9, and 15. Removing these 3 elements from the 11 elements of PQP \cup Q leaves exactly 8 elements.

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1
Identify the elements of sets P, Q, and R within the universal set U.
U={1,2,3,,20}U = \{1, 2, 3, \dots, 20\}, P={2,3,5,7,11,13,17,19}P = \{2, 3, 5, 7, 11, 13, 17, 19\}, Q={1,3,5,7,9,11,13,15,17,19}Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, and R={3,6,9,12,15,18}R = \{3, 6, 9, 12, 15, 18\}.
Listing elements helps accurately compute set unions and intersections.
2
Find the union of sets P and Q, denoted as P ∪ Q.
PQ={1,2,3,5,7,9,11,13,15,17,19}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, so n(PQ)=11n(P \cup Q) = 11.
Combining elements of both P and Q without repetition gives their union.
3
Find the complement of set R with respect to U, denoted as R'.
R={1,2,4,5,7,8,10,11,13,14,16,17,19,20}R' = \{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20\}.
The complement set R' contains all elements in U that are not multiples of 3.
4
Determine the intersection of (P ∪ Q) and R'.
(PQ)R={1,2,5,7,11,13,17,19}(P \cup Q) \cap R' = \{1, 2, 5, 7, 11, 13, 17, 19\}. The number of elements is 8.
This removes the multiples of 3 (namely 3, 9, and 15) from the set PQP \cup Q.

Anahtar Kavram

Set operations including union, intersection, and set complementation.
Tahmini Süre:1m 30s
Soru 4049Soru

Which of the following physical quantities is classified as a fundamental quantity in the International System of Units (SI)?

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Cevap: Electric current

Cevap

Electric current
Electric current is one of the seven fundamental physical quantities defined by the International System of Units (SI). It serves as a base quantity from which other electrical quantities (such as charge and potential difference) are derived.

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1
Identify the definition of a fundamental quantity
Fundamental quantities are independent physical quantities that cannot be expressed in terms of other quantities.
SI defines seven basic quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.
2
Evaluate the options against the list of SI fundamental quantities
Electric current is a basic quantity (measured in amperes, AA), whereas force, velocity, and density are derived from mass, length, and time.
Only electric current is an independent base quantity.

Anahtar Kavram

Fundamental quantities are basic physical quantities that are independent of other quantities.
Soru 4050Soru

Which of the following represents the complete set of real values of xx that satisfy the inequality x2x6x10\frac{x^2 - x - 6}{x - 1} \le 0?

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Cevap: x2x \le -2 or 1<x31 < x \le 3

Cevap

The complete solution set is x2x \le -2 or 1<x31 < x \le 3.
The expression (x3)(x+2)x1\frac{(x-3)(x+2)}{x-1} evaluates to a non-positive value (0\le 0) when the numerator and denominator have opposite signs or when the numerator is zero. Evaluating across the critical boundaries x=2,1,3x = -2, 1, 3 while excluding x=1x = 1 yields the solution set x2x \le -2 or 1<x31 < x \le 3.

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1
Factor the quadratic numerator and state the domain restriction.
The numerator factors as x2x6=(x3)(x+2)x^2 - x - 6 = (x - 3)(x + 2), giving the rational inequality (x3)(x+2)x10\frac{(x - 3)(x + 2)}{x - 1} \le 0 with x1x \neq 1.
Factoring isolates the critical boundary points where the expression can change sign.
2
Identify all critical numbers.
The critical values are x=2x = -2, x=1x = 1, and x=3x = 3.
These points partition the real number line into sub-intervals.
3
Test points in each interval to determine the sign of the rational function f(x)=(x3)(x+2)x1f(x) = \frac{(x - 3)(x + 2)}{x - 1}.
For x<2x < -2, f(x)0f(x) \le 0; for 2<x<1-2 < x < 1, f(x)>0f(x) > 0; for 1<x<31 < x < 3, f(x)0f(x) \le 0; for x>3x > 3, f(x)>0f(x) > 0.
Determining where the function is negative or zero identifies the regions satisfying 0\le 0.
4
Combine intervals and include non-undefined endpoints.
Endpoints x=2x = -2 and x=3x = 3 make the numerator zero (included), while x=1x = 1 makes the denominator zero (excluded). Thus, x2x \le -2 or 1<x31 < x \le 3.
Division by zero must be excluded from the solution set.

Anahtar Kavram

Solving rational inequalities by finding critical values, using test intervals, and respecting domain restrictions.
Soru 4051Soru

Read the sentence below carefully and complete the literary analysis by filling in the missing term. What figure of speech is demonstrated in the sentence?

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In the sentence, 'The exhausted detective spent a sleepless night searching for the missing evidence,' the transfer of the modifier 'sleepless' from the detective to the noun 'night' is an example of .
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Cevap

transferred epithet (or hypallage)
The correct response is 'transferred epithet' (or 'hypallage'). This figure of speech takes place when an epithet or qualifying adjective ('sleepless') is transferred from the person experiencing the condition (the detective) to a related noun ('night').

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1
Analyze the modifier and its modified noun in the sentence.
The adjective 'sleepless' is syntactically attached to the noun 'night'.
Logically, a period of time such as a 'night' cannot experience sleep or sleeplessness; only a living person (the detective) can.
2
Identify the literary device where a modifier is shifted from its logical subject to another associated word.
The device is a transferred epithet (also known in classical rhetoric as hypallage).
A transferred epithet occurs when an adjective (epithet) qualifies a noun other than the person or thing it logically describes.

Anahtar Kavram

Transferred Epithet (Hypallage)
Soru 4052Soru
Find the real value of xx that satisfies the exponential equation 8x+24x1=16x1\frac{8^{x + 2}}{4^{x - 1}} = 16^{x - 1}
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Cevap: 4

Cevap

The value of xx is 44.
Rewriting the terms in base 2 gives 23(x+2)/22(x1)=24(x1)2^{3(x+2)} / 2^{2(x-1)} = 2^{4(x-1)}. Applying the quotient rule gives an exponent of (3x+6)(2x2)=x+8(3x + 6) - (2x - 2) = x + 8 on the left. Equating the exponents yields x+8=4x4x + 8 = 4x - 4, which solves cleanly to x=4x = 4.

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1
Express all terms using a common prime base of 2
The equation becomes (23)x+2(22)x1=(24)x1\frac{(2^3)^{x + 2}}{(2^2)^{x - 1}} = (2^4)^{x - 1}.
Converting to a common base enables the use of index laws to simplify the equation.
2
Apply the power-of-a-power and quotient laws of indices
The left side simplifies to 23(x+2)2(x1)=2x+82^{3(x+2) - 2(x-1)} = 2^{x + 8} and the right side is 24x42^{4x - 4}.
When dividing powers of the same base, exponents are subtracted: am÷an=amna^m \div a^n = a^{m-n}.
3
Equate exponents and solve the linear equation
x+8=4x4    3x=12    x=4x + 8 = 4x - 4 \implies 3x = 12 \implies x = 4.
Because the bases on both sides are identical, their respective exponents must be equal.

Anahtar Kavram

Solving exponential equations using common base conversion and laws of indices
Soru 4053Soru

A physical quantity XX is defined by the expression X=PVmtX = \frac{P \cdot V}{m \cdot t}, where PP represents pressure, VV represents volume, mm represents mass, and tt represents time. Which of the following statements correctly classifies XX and expresses its unit strictly in terms of fundamental SI base units?

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Cevap: XX is a derived quantity, and its unit in SI base units is m2s3\text{m}^2 \cdot \text{s}^{-3}.

Cevap

The physical quantity XX is a derived quantity, and its unit expressed strictly in fundamental SI base units is m2s3\text{m}^2 \cdot \text{s}^{-3}.
The quantity XX is defined via a mathematical formula involving pressure, volume, mass, and time, which classifies it as a derived quantity. Replacing each component with its fundamental SI base units gives pressure as kgm1s2\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} and volume as m3\text{m}^3, making the numerator kgm2s2\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}. Dividing by the denominator (mt=kgsm \cdot t = \text{kg} \cdot \text{s}) cancels out kilograms and leaves m2s3\text{m}^2 \cdot \text{s}^{-3}, consisting purely of fundamental base units.

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1
Classify the physical quantity XX
XX is a derived physical quantity.
Fundamental physical quantities in the SI system are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Because XX is calculated from a combination of other quantities, it is derived.
2
Express pressure (PP) and volume (VV) in terms of fundamental SI base units
P=kgm1s2P = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} and V=m3V = \text{m}^3.
Pressure is defined as force per unit area (kgms2m2)\left(\frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}^2}\right), and volume has the base unit m3\text{m}^3.
3
Calculate the base unit expression for the numerator PVP \cdot V
PV=(kgm1s2)(m3)=kgm2s2P \cdot V = (\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}) \cdot (\text{m}^3) = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}.
Combining the powers of length (metres) gives 1+3=2-1 + 3 = 2.
4
Divide by the denominator mtm \cdot t to obtain the base units of XX
X=kgm2s2kgs=m2s3X = \frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{kg} \cdot \text{s}} = \text{m}^2 \cdot \text{s}^{-3}.
The unit of mass (kg\text{kg}) cancels completely, and dividing by time (s\text{s}) reduces the exponent of seconds from 2-2 to 3-3.

Anahtar Kavram

Fundamental quantities are independent base quantities defined by the SI system, whereas derived quantities are defined algebraically from fundamental quantities. Reducing derived units to SI base units requires breaking down all non-base units into metres (m), kilograms (kg), seconds (s), amperes (A), kelvins (K), moles (mol), or candelas (cd).
Tahmini Süre:2m 0s
Soru 4054Soru

A copper calorimeter of heat capacity 300 J K1300\text{ J K}^{-1} contains 0.5 kg0.5\text{ kg} of water at an initial temperature of 25C25^\circ\text{C}. An electric heater rated at 800 W800\text{ W} is immersed in the water to heat the system for 4 minutes4\text{ minutes}. If heat is lost to the surrounding environment at a constant rate of 200 W200\text{ W} throughout the heating duration, what is the final temperature of the water-calorimeter system in C^\circ\text{C}? (Take the specific heat capacity of water as 4200 J kg1K14200\text{ J kg}^{-1}\text{K}^{-1})

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Cevap: 85

Cevap

The final temperature of the system is 85C85^\circ\text{C}.
The net thermal energy added to the system accounts for both the supplied electrical energy and the heat energy lost to the surroundings: Qnet=(800200) W×240 s=144,000 JQ_{\text{net}} = (800 - 200)\text{ W} \times 240\text{ s} = 144,000\text{ J}. The total heat capacity of the water-calorimeter system is Ctotal=300 J K1+(0.5 kg×4200 J kg1K1)=2400 J K1C_{\text{total}} = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}. The temperature increase is ΔT=144,0002400=60C\Delta T = \frac{144,000}{2400} = 60^\circ\text{C}. Adding this to the initial temperature of 25C25^\circ\text{C} gives the final temperature of 85C85^\circ\text{C}.

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1
Convert heating time to standard SI units (seconds)
t=4×60 s=240 st = 4 \times 60\text{ s} = 240\text{ s}
Power is measured in Joules per second (Watts), so time must be in seconds.
2
Calculate the net rate of heat energy input to the system
Pnet=800 W200 W=600 WP_{\text{net}} = 800\text{ W} - 200\text{ W} = 600\text{ W}
The net heating power is the input power minus the power dissipated as heat loss.
3
Calculate total net heat energy transferred to the system
Qnet=Pnet×t=600 W×240 s=144,000 JQ_{\text{net}} = P_{\text{net}} \times t = 600\text{ W} \times 240\text{ s} = 144,000\text{ J}
Thermal energy transferred equals net power multiplied by time.
4
Compute the total heat capacity of the combined system (water + calorimeter)
Ctotal=Ccalorimeter+(mwater×cwater)=300 J K1+(0.5 kg×4200 J kg1K1)=2400 J K1C_{\text{total}} = C_{\text{calorimeter}} + (m_{\text{water}} \times c_{\text{water}}) = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}
Heat capacity of water is mass multiplied by specific heat capacity, added to the calorimeter's given heat capacity.
5
Calculate the temperature rise of the system
ΔT=QnetCtotal=144,000 J2400 J K1=60C\Delta T = \frac{Q_{\text{net}}}{C_{\text{total}}} = \frac{144,000\text{ J}}{2400\text{ J K}^{-1}} = 60^\circ\text{C}
Temperature change is total net heat supplied divided by total heat capacity.
6
Find the final temperature of the system
Tfinal=Tinitial+ΔT=25C+60C=85CT_{\text{final}} = T_{\text{initial}} + \Delta T = 25^\circ\text{C} + 60^\circ\text{C} = 85^\circ\text{C}
Final temperature equals initial temperature plus temperature increase.

Anahtar Kavram

Conservation of thermal energy in calorimeter systems with continuous power loss
Tahmini Süre:1m 30s
Soru 4055Soru

A factory uses 88 identical machines to produce 480480 items in 66 hours. If 33 of the machines break down, how many hours will it take the remaining machines to produce 600600 items at the same rate?

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Cevap: 12

Cevap

The remaining machines will take 12 hours to produce 600 items.
Each machine produces 4808×6=10\frac{480}{8 \times 6} = 10 items per hour. With 55 active machines, the combined rate is 5050 items per hour. To reach 600600 items, the time required is 60050=12\frac{600}{50} = 12 hours.

Adım Adım Çözüm

1
Calculate the output rate per machine per hour
10 items per machine per hour
The combined rate of 8 machines is 480÷6=80480 \div 6 = 80 items per hour. Dividing by 8 machines gives 1010 items/hour per machine.
2
Determine the new combined production rate
50 items per hour
With 3 machines out of service, 83=58 - 3 = 5 machines remain, working at a total rate of 5×10=505 \times 10 = 50 items per hour.
3
Calculate total hours required for the new target quantity
12 hours
Dividing the target quantity of 600600 items by the rate of 5050 items/hour yields 60050=12\frac{600}{50} = 12 hours.

Anahtar Kavram

Compound Proportion and Work Rate
Tahmini Süre:1m 30s
Soru 4056Soru

An electron of mass 9.1×1031 kg9.1 \times 10^{-31}\text{ kg} and charge 1.6×1019 C1.6 \times 10^{-19}\text{ C} enters perpendicularly into a uniform magnetic field of flux density 2.0×103 T2.0 \times 10^{-3}\text{ T} with a speed of 3.2×106 m/s3.2 \times 10^6\text{ m/s}. What is the radius of the circular path followed by the electron, expressed in millimeters (mm)?

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Cevap: 9.1

Cevap

The radius of the circular path followed by the electron is 9.1 mm.
When a charge qq enters a magnetic field BB perpendicularly at speed vv, the magnetic force qvBqvB supplies the centripetal force mv2r\frac{mv^2}{r}. Rearranging for radius yields r=mvqBr = \frac{mv}{qB}. Substituting m=9.1×1031 kgm = 9.1 \times 10^{-31}\text{ kg}, v=3.2×106 m/sv = 3.2 \times 10^6\text{ m/s}, q=1.6×1019 Cq = 1.6 \times 10^{-19}\text{ C}, and B=2.0×103 TB = 2.0 \times 10^{-3}\text{ T} gives r=9.1×103 m=9.1 mmr = 9.1 \times 10^{-3}\text{ m} = 9.1\text{ mm}.

Adım Adım Çözüm

1
Equate the magnetic force to the centripetal force for circular motion
qvB=mv2rqvB = \frac{mv^2}{r}
A charged particle moving perpendicularly to a magnetic field experiences a magnetic force that acts entirely as a centripetal force.
2
Rearrange the equation to make the orbital radius rr the subject
r=mvqBr = \frac{mv}{qB}
Cancelling one factor of velocity vv from both sides allows direct computation of rr.
3
Substitute the physical values into the formula
r=(9.1×1031)(3.2×106)(1.6×1019)(2.0×103)=9.1×103 mr = \frac{(9.1 \times 10^{-31})(3.2 \times 10^6)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})} = 9.1 \times 10^{-3}\text{ m}
Calculates the radius in standard SI units (meters).
4
Convert the resulting radius from meters to millimeters
r=9.1×103 m×1000 mm/m=9.1 mmr = 9.1 \times 10^{-3}\text{ m} \times 1000\text{ mm/m} = 9.1\text{ mm}
The question explicitly requests the answer in millimeters.

Anahtar Kavram

Motion of a charged particle in a uniform magnetic field
Soru 4057Soru

A gas cylinder fitted with a frictionless piston contains a fixed mass of ideal gas occupying a volume of 0.040 m30.040\text{ m}^3 at a pressure of 1.50×105 Pa1.50 \times 10^5\text{ Pa} and a temperature of 27C27^\circ\text{C}. The gas is heated to 127C127^\circ\text{C} while expanding to a new volume of 0.080 m30.080\text{ m}^3. What is the final pressure of the gas?

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Cevap: 1.00×105 Pa1.00 \times 10^5\text{ Pa}

Cevap

The final pressure of the gas is 1.00×105 Pa1.00 \times 10^5\text{ Pa}.
By converting temperatures to absolute zero scale (300 K300\text{ K} and 400 K400\text{ K}) and applying the Combined Gas Law P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}, the final pressure P2P_2 evaluates directly to 1.00×105 Pa1.00 \times 10^5\text{ Pa}.

Adım Adım Çözüm

1
Convert temperatures from Celsius to Kelvin.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=127+273=400 KT_2 = 127 + 273 = 400\text{ K}.
Gas laws require absolute temperatures measured on the Kelvin scale.
2
Apply the Combined Gas Law formula P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}.
(1.50×105 Pa)(0.040 m3)300 K=P2(0.080 m3)400 K\frac{(1.50 \times 10^5\text{ Pa})(0.040\text{ m}^3)}{300\text{ K}} = \frac{P_2 (0.080\text{ m}^3)}{400\text{ K}}.
The mass of the gas is fixed while pressure, volume, and temperature all change.
3
Rearrange and solve for final pressure P2P_2.
P2=1.50×105×(0.0400.080)×(400300)=1.00×105 PaP_2 = 1.50 \times 10^5 \times \left(\frac{0.040}{0.080}\right) \times \left(\frac{400}{300}\right) = 1.00 \times 10^5\text{ Pa}.
Simplifying the numerical expression yields the final equilibrium pressure.

Anahtar Kavram

Combined Gas Law
Tahmini Süre:2m 0s
Soru 4058Soru

Match each derived SI unit listed on the left with its corresponding fundamental (base) SI unit representation on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Joule (J\text{J})
Pascal (Pa\text{Pa})
Watt (W\text{W})
Newton (N\text{N})

Eşleşmeler

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Cevap

Joule (J\text{J}) matches kgm2s2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}; Pascal (Pa\text{Pa}) matches kgm1s2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}; Watt (W\text{W}) matches kgm2s3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}; Newton (N\text{N}) matches kgms2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}.
Each derived unit is systematically expressed in terms of the fundamental SI units of mass (kg), length (m), and time (s) by substituting definitions: Newton is kgms2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}, Joule is kgm2s2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}, Pascal is kgm1s2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, and Watt is kgm2s3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}.

Adım Adım Çözüm

1
Express Newton (N) in fundamental units
Force=mass×acceleration1 N=1 kgms2\text{Force} = \text{mass} \times \text{acceleration} \Rightarrow 1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}
Force is defined as mass multiplied by acceleration.
2
Express Joule (J) in fundamental units
Work=Force×distance1 J=(kgms2)×m=1 kgm2s2\text{Work} = \text{Force} \times \text{distance} \Rightarrow 1\text{ J} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}
Work done is force multiplied by displacement in the direction of the force.
3
Express Pascal (Pa) in fundamental units
Pressure=ForceArea1 Pa=kgms2m2=1 kgm1s2\text{Pressure} = \frac{\text{Force}}{\text{Area}} \Rightarrow 1\text{ Pa} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Pressure is defined as force applied perpendicular to a surface per unit area.
4
Express Watt (W) in fundamental units
Power=Worktime1 W=kgm2s2s=1 kgm2s3\text{Power} = \frac{\text{Work}}{\text{time}} \Rightarrow 1\text{ W} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{s}} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-3}
Power is the rate at which work is done or energy is transferred.

Anahtar Kavram

Expressing derived SI units in terms of base SI fundamental units
Soru 4059Soru

Which qualitative characteristic of accounting information ensures that financial statements contain data capable of making a difference in the economic decisions of users?

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Cevap: Relevance

Cevap

Relevance is the primary qualitative characteristic that makes accounting information capable of influencing user decisions.
Accounting information possesses relevance when it is capable of making a difference in the decisions made by users. It achieves this by helping users evaluate past, present, or future events (predictive value) or confirming/correcting previous expectations (confirmatory value).

Adım Adım Çözüm

1
Identify the core requirement of the question.
The question asks for the qualitative characteristic that directly makes financial information capable of making a difference in decision-making.
Understanding the definition of fundamental qualitative characteristics is key to selecting the correct accounting attribute.
2
Distinguish between fundamental and enhancing qualitative characteristics.
Fundamental characteristics are Relevance and Faithful Representation. Enhancing characteristics are Comparability, Verifiability, Timeliness, and Understandability.
Relevance directly addresses decision-usefulness through predictive and confirmatory value.

Anahtar Kavram

Fundamental Qualitative Characteristics of Accounting Information
Soru 4060Soru

The following trial balance extract was taken from the books of a sole proprietor as at 31st December 2025:

Ledger AccountAmount (\text{₦})
Sales200,000
Returns inwards10,000
Purchases130,000
Returns outwards6,000
Opening inventory30,000
Carriage inwards4,000
Carriage outwards8,000

Additional Information:
1. Goods costing 4,000\text{₦}4,000 were withdrawn by the proprietor for personal use during the financial year, but no entry has been made in the accounting records.
2. The business sells goods at a uniform mark-up of 25%25\% on cost of goods sold.

What is the value of closing inventory to be recorded in the trading account?

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Cevap: ₦2,000

Cevap

The value of closing inventory to be recorded in the trading account is ₦2,000.
Net sales equal ₦190,000 (Gross sales ₦200,000 less Returns inwards ₦10,000). Convert the 25% mark-up (14\frac{1}{4} on cost) to margin (15\frac{1}{5} or 20% on sales), yielding Gross Profit of ₦38,000 (20%×190,00020\% \times \text{₦}190,000) and Cost of Goods Sold (COGS) of ₦152,000 (190,00038,000\text{₦}190,000 - \text{₦}38,000). Cost of goods available for sale is ₦154,000 (Opening stock ₦30,000 + Purchases ₦130,000 - Returns outwards ₦6,000 - Goods withdrawn ₦4,000 + Carriage inwards ₦4,000). Subtracting COGS (₦152,000) from Cost of Goods Available for Sale (₦154,000) gives a closing inventory of ₦2,000.

Adım Adım Çözüm

1
Calculate Net Sales
Net Sales = ₦200,000 - ₦10,000 = ₦190,000
Returns inwards must be deducted from gross sales to obtain net sales revenue.
2
Convert Mark-up to Margin and compute Gross Profit and Cost of Goods Sold (COGS)
Margin = 20%; Gross Profit = ₦38,000; COGS = ₦152,000
A mark-up of 25%25\% (14\frac{1}{4}) on cost corresponds to a margin of 11+4=15=20%\frac{1}{1 + 4} = \frac{1}{5} = 20\% on net sales. Gross Profit = 20%×190,000=38,00020\% \times \text{₦}190,000 = \text{₦}38,000. COGS = Net Sales - Gross Profit = 190,00038,000=152,000\text{₦}190,000 - \text{₦}38,000 = \text{₦}152,000.
3
Calculate Adjusted Net Purchases
Adjusted Net Purchases = ₦130,000 - ₦6,000 - ₦4,000 = ₦120,000
Returns outwards and owner's inventory drawings must be deducted from gross purchases.
4
Calculate Cost of Goods Available for Sale
Goods Available = ₦30,000 + ₦120,000 + ₦4,000 = ₦154,000
Carriage inwards is added to opening inventory and net purchases as a direct cost of getting goods into the business. Carriage outwards is a selling expense and is excluded.
5
Determine Closing Inventory
Closing Inventory = ₦154,000 - ₦152,000 = ₦2,000
Closing Inventory = Cost of Goods Available for Sale - Cost of Goods Sold.

Anahtar Kavram

Trading Account, COGS Determination, and Mark-up/Margin Conversion
Tahmini Süre:3m 0s
ÖncekiSayfa 203 / 697Sonraki
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