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Question 7561Question

The critical velocity vcv_c of a fluid flowing through a cylindrical pipe of diameter DD depends on the dynamic viscosity η\eta of the fluid, its density ρ\rho, and the pipe diameter DD according to the empirical relationship vc=ReηxρyDzv_c = R_e \eta^x \rho^y D^z, where ReR_e is the dimensionless Reynolds number. Using dimensional analysis, calculate the numerical value of the sum of the exponents x+y+zx + y + z.

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Answer: -1

Answer

The sum of the exponents x+y+zx + y + z is 1-1.
Using the principle of dimensional homogeneity, the dimensions of both sides of the formula vc=ReηxρyDzv_c = R_e \eta^x \rho^y D^z must be equal. Equating the powers of Mass, Length, and Time yields x=1x = 1, y=1y = -1, and z=1z = -1. Summing these three values gives 1+(1)+(1)=11 + (-1) + (-1) = -1.

Step-by-Step Solution

1
Determine the dimensions of all physical quantities involved
Critical velocity [vc]=L T1[v_c] = \text{L T}^{-1}, dynamic viscosity [η]=M L1T1[\eta] = \text{M L}^{-1} \text{T}^{-1}, fluid density [ρ]=M L3[\rho] = \text{M L}^{-3}, and diameter [D]=L[D] = \text{L}. The Reynolds number ReR_e is dimensionless.
Dimensional analysis requires replacing physical quantities with their base SI dimensions.
2
Formulate the dimensional homogeneity equation
\text{M}^0 \text{L}^1 \text{T}^{-1} = (\text{M L}^{-1} \text{T}^{-1})^x (\text{M L}^{-3})^y (\text{L})^z = \text{M}^{x+y} \text{L}^{-x-3y+z} \text{T}^{-x}.
By the principle of dimensional homogeneity, the total exponent of each fundamental dimension must match on both sides of the equation.
3
Solve the system of simultaneous linear equations for xx, yy, and zz
From T\text{T}: x=1    x=1-x = -1 \implies x = 1.
From M\text{M}: x+y=0    y=1x + y = 0 \implies y = -1.
From L\text{L}: x3y+z=1    1+3+z=1    z=1-x - 3y + z = 1 \implies -1 + 3 + z = 1 \implies z = -1.
Equating powers of fundamental quantities yields explicit values for each dimensional power.
4
Calculate the target sum x+y+zx + y + z
x + y + z = 1 + (-1) + (-1) = -1.
Combining the calculated exponents gives the required numerical value.

Key Concept

Principle of Dimensional Homogeneity and Derivation of Physical Formulas
Question 7562Question

Given the simultaneous equations 2xy=42x - y = 4 and x2+y2=13x^2 + y^2 = 13, where both xx and yy are positive real numbers, calculate the value of x+yx + y.

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Answer: 5

Answer

The correct value of x+yx + y is 5.
Substituting y=2x4y = 2x - 4 into x2+y2=13x^2 + y^2 = 13 gives 5x216x+3=05x^2 - 16x + 3 = 0, which yields x=3x = 3 or x=0.2x = 0.2. The corresponding yy-values are y=2y = 2 and y=3.6y = -3.6. Since both xx and yy must be positive, the valid pair is (3,2)(3, 2), giving x+y=3+2=5x + y = 3 + 2 = 5.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation 2xy=42x - y = 4.
y=2x4y = 2x - 4
Substitution is the standard method for solving simultaneous linear and quadratic equations.
2
Substitute y=2x4y = 2x - 4 into the quadratic equation x2+y2=13x^2 + y^2 = 13.
x2+(2x4)2=13x^2 + (2x - 4)^2 = 13
This reduces the system to a single quadratic equation in one variable.
3
Expand and collect like terms.
5x216x+3=05x^2 - 16x + 3 = 0
Expanding (2x4)2=4x216x+16(2x - 4)^2 = 4x^2 - 16x + 16 and subtracting 13 puts the quadratic in standard form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation 5x216x+3=05x^2 - 16x + 3 = 0 for xx.
x=3x = 3 or x=0.2x = 0.2
Factoring (5x1)(x3)=0(5x - 1)(x - 3) = 0 yields two real solutions for xx.
5
Find corresponding values of yy and apply the positivity constraint x>0x > 0 and y>0y > 0.
x=3,y=2x = 3, y = 2
When x=0.2x = 0.2, y=3.6y = -3.6, which is not positive. Hence, (3,2)(3, 2) is the only valid solution pair.
6
Calculate x+yx + y.
5
3+2=53 + 2 = 5.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Estimated Time:1m 30s
Question 7563Question

A curve has a gradient function defined by dydx=12x36sin(3x)+4\frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4. If the curve passes through the point (0,15)(0, 15), what is the value of the constant of integration CC?

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Answer: 13

Answer

The constant of integration CC is 13.
Integrating 12x36sin(3x)+412x^3 - 6\sin(3x) + 4 with respect to xx yields y=3x4+2cos(3x)+4x+Cy = 3x^4 + 2\cos(3x) + 4x + C. Substituting x=0x = 0 and y=15y = 15 gives 15=3(0)4+2cos(0)+4(0)+C=2+C15 = 3(0)^4 + 2\cos(0) + 4(0) + C = 2 + C, which simplifies to C=13C = 13.

Step-by-Step Solution

1
Integrate the gradient function to obtain the general equation of the curve.
y=3x4+2cos(3x)+4x+Cy = 3x^4 + 2\cos(3x) + 4x + C
The integral of 12x312x^3 is 3x43x^4, the integral of 6sin(3x)-6\sin(3x) is +2cos(3x)+2\cos(3x), and the integral of 44 is 4x4x.
2
Apply the initial condition (0,15)(0, 15) to solve for CC.
15=3(0)4+2cos(0)+4(0)+C15 = 3(0)^4 + 2\cos(0) + 4(0) + C
Substitute x=0x = 0 and y=15y = 15 into the integrated equation.
3
Calculate the numerical value of CC.
C=13C = 13
Since cos(0)=1\cos(0) = 1, the equation becomes 15=2+C15 = 2 + C, yielding C=13C = 13.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions
Question 7564Question

Given the matrices A=(x21302141)A = \begin{pmatrix} x & 2 & 1 \\ 3 & 0 & 2 \\ 1 & 4 & 1 \end{pmatrix} and B=(2153)B = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}, find the value of xx if det(A)+det(B)=5\det(A) + \det(B) = 5.

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Answer: 2

Answer

The value of x is 2.
Evaluating \det(B) gives 11, and cofactor expansion of \det(A) along the top row yields 10 - 8x. Summing these determinants gives (10 - 8x) + 11 = 5, which simplifies to 8x = 16, resulting in x = 2.

Step-by-Step Solution

1
Calculate the determinant of the 2x2 matrix B
\det(B) = 11
For a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad - bc. Thus, \det(B) = (2)(3) - (-1)(5) = 6 + 5 = 11.
2
Calculate the determinant of the 3x3 matrix A in terms of x
\det(A) = 10 - 8x
Expanding along the first row: \det(A) = x\begin{vmatrix} 0 & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 0 \\ 1 & 4 \end{vmatrix} = x(0 - 8) - 2(3 - 2) + 1(12 - 0) = -8x - 2 + 12 = 10 - 8x.
3
Set up and solve the linear equation \det(A) + \det(B) = 5
x = 2
Substituting the computed determinants gives (10 - 8x) + 11 = 5, which simplifies to 21 - 8x = 5. Rearranging yields 8x = 16, so x = 2.

Key Concept

Determinants of 2x2 and 3x3 matrices and linear equation solving
Question 7565Question

A business owner borrowed 150,000\text{₦}150,000 to expand his store at a compound interest rate of 8%8\% per annum, compounded annually. What is the total compound interest he will pay at the end of 2 years?

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Answer: 24,960\text{₦}24,960

Answer

The total compound interest paid at the end of 2 years is 24,960\text{₦}24,960.
The total compound interest is obtained by calculating the interest accrued in each period (12,00012,000 in Year 1 and 12,96012,960 in Year 2) and summing them to get 24,960\text{₦}24,960. Alternatively, using the compound interest formula I=P(1+r)nPI = P(1 + r)^n - P, I=150,000(1.08)2150,000=174,960150,000=24,960I = 150,000(1.08)^2 - 150,000 = 174,960 - 150,000 = \text{₦}24,960.

Step-by-Step Solution

1
Calculate the interest for the first year.
Interest1=8100×150,000=12,000\text{Interest}_1 = \frac{8}{100} \times 150,000 = \text{₦}12,000
Interest in the first year is calculated on the initial principal.
2
Find the principal for the second year.
Principal2=150,000+12,000=162,000\text{Principal}_2 = 150,000 + 12,000 = \text{₦}162,000
Under compound interest, the first year's interest is added to the initial principal.
3
Calculate the interest for the second year.
Interest2=8100×162,000=12,960\text{Interest}_2 = \frac{8}{100} \times 162,000 = \text{₦}12,960
Interest in the second year is calculated on the updated principal.
4
Sum the interest from both years to find the total compound interest.
Total Interest=12,000+12,960=24,960\text{Total Interest} = 12,000 + 12,960 = \text{₦}24,960
The total compound interest is the sum of interest accumulated in each compounding period.

Key Concept

Compound Interest
Estimated Time:1m 30s
Question 7566Question

During Rutherford's alpha particle scattering experiment, the vast majority of alpha particles passed straight through the gold foil with little to no deflection. What conclusion about atomic structure was drawn directly from this specific observation?

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Answer: The atom consists mostly of empty space

Answer

The atom consists mostly of empty space
In Rutherford's alpha particle scattering experiment, the fact that over 99% of alpha particles passed straight through the foil undeflected indicates that they encountered no massive positive obstacle. This directly proved that the volume of an atom consists predominantly of empty space, with positive charge concentrated in a tiny central region called the nucleus.

Step-by-Step Solution

1
Analyze the experimental observation from Rutherford's experiment
The vast majority of positively charged alpha particles suffered no deflection when passing through gold foil.
Deflection occurs when alpha particles pass close to a dense positive charge.
2
Deduce the physical implication for the structure of the atom
Since almost no deflection occurred for most particles, they encountered no heavy concentration of positive charge or mass in their path.
This proves that the dense positively charged nucleus occupies an extremely small fraction of the total atomic volume, leaving the rest as empty space.

Key Concept

Rutherford's Alpha Scattering Experiment and Nuclear Atom Model
Question 7567Question

A signal generator positioned inside a transparent, sealed glass chamber produces both sound waves and radio waves simultaneously. As an air pump gradually evacuates the chamber to create a vacuum, which of the following best describes what happens to the propagation of these two waves?

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Answer: Sound waves cease to propagate, whereas radio waves continue to propagate through the chamber.

Answer

Sound waves cease to propagate because they are mechanical waves requiring a physical medium, whereas radio waves continue to propagate because they are electromagnetic waves capable of traveling through a vacuum.
Sound waves are mechanical waves that rely on the vibration of medium particles (such as air molecules) to propagate energy. In contrast, radio waves are electromagnetic waves composed of mutually perpendicular oscillating electric and magnetic fields, allowing them to propagate efficiently through both matter and vacuum. Thus, evacuating air stops sound transmission while radio transmission continues.

Step-by-Step Solution

1
Classify the wave types produced by the generator
Sound waves are mechanical waves (longitudinal), while radio waves are electromagnetic waves (transverse).
Wave propagation mechanisms depend fundamental on whether a wave is mechanical or electromagnetic.
2
Determine medium requirements for each wave class
Mechanical waves require elastic material particles to transfer vibrational energy; electromagnetic waves consist of oscillating electric and magnetic fields which travel without a physical medium.
In a vacuum, particle density approaches zero, preventing mechanical wave transmission.
3
Evaluate the effect of evacuating the glass chamber
As air is removed, sound wave propagation dies out, whereas radio wave propagation remains unimpeded.
The absence of matter blocks sound propagation while permitting electromagnetic wave transmission.

Key Concept

Classification of Waves by Medium Requirement (Mechanical vs. Electromagnetic Waves)
Question 7568Question

A research committee of 55 members is to be formed from 66 male scientists and 55 female scientists. The committee must contain strictly more male scientists than female scientists. If one particular male scientist and one particular female scientist refuse to serve on the same committee together, in how many different ways can the committee be selected?

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Answer: 231

Answer

The total number of valid ways to select the committee is 231.
The correct response of 231 is found by calculating the total committees having strictly more men than women (281) and subtracting the subset of those committees where both restricted individuals are chosen together (50).

Step-by-Step Solution

1
Calculate the unconstrained ways to form a 5-member committee with strictly more males than females.
Three possible gender distributions exist:
- 5 Males, 0 Females: (65)×(50)=6×1=6\binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6
- 4 Males, 1 Female: (64)×(51)=15×5=75\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75
- 3 Males, 2 Females: (63)×(52)=20×10=200\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200
Total unconstrained ways = 6+75+200=2816 + 75 + 200 = 281.
Establishing the total pool of valid gender-ratio committees before applying the exclusion condition.
2
Calculate the number of forbidden committee selections where both the restricted male scientist and restricted female scientist are included together.
Since 1 male and 1 female are already chosen, 3 remaining members must be selected from the remaining 5 males and 4 females:
- For 4 Males, 1 Female total (need 3 more males, 0 more females): (53)×(40)=10×1=10\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10
- For 3 Males, 2 Females total (need 2 more males, 1 more female): (52)×(41)=10×4=40\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40
Total forbidden ways = 10+40=5010 + 40 = 50.
To enforce the restriction that both specific members cannot serve together, calculate all committee configurations containing both.
3
Subtract the forbidden selections from the unconstrained selections.
Valid ways = 28150=231281 - 50 = 231.
Applying the subtraction principle of combinations to yield the final allowed arrangements.

Key Concept

Combinations with gender ratio constraints and mutual exclusion restrictions
Estimated Time:2m 30s
Question 7569Question

A binary operation \star is defined on the set of real numbers R\mathbb{R} by ab=a+b+7a \star b = a + b + 7. What is the identity element of the operation?

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Answer: 7-7

Answer

The identity element of the operation is 7-7.
For an identity element ee, the condition ae=aa \star e = a must hold for all real numbers aa. Substituting the operation definition gives a+e+7=aa + e + 7 = a. Subtracting aa from both sides results in e+7=0e + 7 = 0, which solves to e=7e = -7.

Step-by-Step Solution

1
Set up the defining equation for the identity element ee.
ae=aa \star e = a
By definition, an identity element ee leaves any element aa unchanged under the operation.
2
Apply the given rule for the binary operation.
a+e+7=aa + e + 7 = a
The binary operation is defined as ab=a+b+7a \star b = a + b + 7.
3
Solve for ee.
e=7e = -7
Subtracting a+7a + 7 from both sides gives e=7e = -7.

Key Concept

Identity element of a binary operation
Estimated Time:45s
Question 7570Question

Match each historical atomic model on the left with its defining postulate, experimental outcome, or theoretical limitation on the right.

Click a left item, then click its matching right item

Items

Thomson's Plum Pudding Model
Rutherford's Planetary Model
Bohr's Quantized Model
Sommerfeld's Extension

Matches

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Answer

Thomson's model matches the diffuse positive sphere disproved by alpha particle backscattering; Rutherford's model matches the dense nucleus with classical radiation collapse limitations; Bohr's model matches quantized angular momentum in non-radiating orbits; Sommerfeld's extension matches elliptical sub-shells and relativistic adjustments for fine-structure splitting.
Each model directly maps to its defining theoretical contribution or failure mechanism: Thomson's diffuse charge sphere failed under α\alpha-particle scattering; Rutherford's nuclear atom suffered from classical radiation instability; Bohr's model introduced angular momentum quantization L=nL = n\hbar; and Sommerfeld's model extended orbits to ellipses with relativistic velocity corrections to account for fine structure.

Step-by-Step Solution

1
Analyze Thomson's Plum Pudding Model
Thomson proposed electrons embedded in a sea of positive charge. This continuous distribution could not account for α\alpha-particles rebounding at angles greater than 9090^\circ.
Identify the historical assumption and experimental contradiction for Thomson's model.
2
Analyze Rutherford's Planetary Model
Rutherford deduced a concentrated positive core (nucleus). However, according to Maxwellian electrodynamics, orbiting electrons accelerate continuously, radiating energy until collapsing into the nucleus.
Identify the primary theoretical failure of classical planetary electron orbits.
3
Analyze Bohr's Quantized Model
Bohr introduced the non-classical postulate that electrons exist in stable stationary states with angular momentum L=nh2πL = \frac{nh}{2\pi}, accurately yielding the Rydberg formula for hydrogen.
Recognize the quantum postulate resolving Rutherford's radiation collapse.
4
Analyze Sommerfeld's Extension
To explain fine spectral line splitting not accounted for by circular Bohr orbits, Sommerfeld introduced elliptical orbits with azimuthal quantum numbers and relativistic mass variation at high electron velocities.
Connect fine-structure spectral features to relativistic elliptical orbital modifications.

Key Concept

Development and Limitations of Historical Atomic Models
Question 7571Question

A committee of 66 delegates is to be seated around a circular table for a conference. If two specific delegates must always sit next to each other, in how many distinct ways can the delegates be arranged?

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Answer: 48

Answer

48
The correct answer is 48 because treating the 2 restricted delegates as a single unit leaves 5 items to arrange around a circle, which yields (51)!=24(5-1)! = 24 arrangements. Since the 2 delegates can arrange themselves in 2!=22! = 2 ways within their block, the total number of arrangements is 24×2=4824 \times 2 = 48.

Step-by-Step Solution

1
Group the restricted delegates into a single block
2 specified delegates are treated as 1 unit, leaving 4 remaining delegates, making a total of 5 items to arrange.
Since the two delegates must sit next to each other, treating them as a single entity ensures they remain together in all seating positions.
2
Calculate the circular arrangements of the 5 items
(51)!=4!=24(5 - 1)! = 4! = 24 ways.
The number of ways to arrange nn distinct items around a circular table is given by (n1)!(n - 1)!.
3
Account for internal arrangements of the paired delegates
2!=22! = 2 ways.
The two delegates within the single block can swap positions between themselves.
4
Multiply the circular arrangements by the internal arrangements
24×2=4824 \times 2 = 48 total distinct arrangements.
By the fundamental counting principle, total arrangements equal the product of the independent steps.

Key Concept

Circular Permutations with Restrictions
Estimated Time:1m 15s
Question 7572Question

Match each physical quantity to its correct classification as either a fundamental or a derived physical quantity.

Click a left item, then click its matching right item

Items

Luminous intensity
Mass
Force
Electric potential

Matches

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Answer

Luminous intensity matches with fundamental quantity measuring light brightness; Mass matches with fundamental quantity measuring quantity of matter; Force matches with derived quantity defined as rate of change of linear momentum; Electric potential matches with derived quantity defined as work done per unit electric charge.
Luminous intensity and mass are two of the seven base SI quantities. Force and electric potential are derived quantities defined through mathematical combinations of base quantities.

Step-by-Step Solution

1
Identify the fundamental physical quantities
Luminous intensity and Mass are fundamental physical quantities.
Fundamental physical quantities are basic quantities that do not depend on any other physical quantity for their definition.
2
Identify the derived physical quantities
Force and Electric potential are derived physical quantities.
Derived physical quantities are obtained by combining fundamental physical quantities through mathematical relationships.

Key Concept

Fundamental quantities (length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity) are independent, whereas derived quantities are formed by combining fundamental quantities.
Question 7573Question

Let the universal set be U={xZ:1x20}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 20\}. Consider two subsets of U\mathcal{U} given by A={xU:x is a multiple of 4}A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\} and B={xU:x is a factor of 20}B = \{x \in \mathcal{U} : x \text{ is a factor of } 20\}. What is the number of elements in the complement of (AB)(A \cup B), denoted as n((AB))n((A \cup B)')?

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Answer: 11

Answer

11
The universal set contains 20 elements. The set of multiples of 4 within this range has 5 elements, and the set of factors of 20 has 6 elements. Their shared elements are 4 and 20 (2 elements). Thus, the union contains 5+62=95 + 6 - 2 = 9 elements. Subtracting this from 20 gives 11 elements in the complement.

Step-by-Step Solution

1
Identify the elements of the universal set and the subsets AA and BB.
U={1,2,3,,20}\mathcal{U} = \{1, 2, 3, \dots, 20\} with n(U)=20n(\mathcal{U}) = 20.
A={4,8,12,16,20}A = \{4, 8, 12, 16, 20\} with n(A)=5n(A) = 5.
B={1,2,4,5,10,20}B = \{1, 2, 4, 5, 10, 20\} with n(B)=6n(B) = 6.
Listing the elements clearly determines the cardinality of each individual set.
2
Find the intersection and union of sets AA and BB.
AB={4,20}A \cap B = \{4, 20\}, so n(AB)=2n(A \cap B) = 2.
AB={1,2,4,5,8,10,12,16,20}A \cup B = \{1, 2, 4, 5, 8, 10, 12, 16, 20\}, so n(AB)=5+62=9n(A \cup B) = 5 + 6 - 2 = 9.
Applying the principle of inclusion-exclusion avoids double-counting common elements.
3
Calculate the cardinality of the complement (AB)(A \cup B)'.
n((AB))=n(U)n(AB)=209=11n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 20 - 9 = 11.
The complement contains all elements in the universal set that are not in the union.

Key Concept

Set Complements and Inclusion-Exclusion Principle
Question 7574Question

A small charged sphere of mass 2.0×104 kg2.0 \times 10^{-4}\text{ kg} carrying a positive charge of +4.0×108 C+4.0 \times 10^{-8}\text{ C} is suspended by a light insulating string between two vertical parallel plates. When a uniform horizontal electric field of magnitude EE is applied between the plates, the string deflects and comes to equilibrium at an angle of 4545^\circ to the vertical. Taking the acceleration due to gravity g=10 ms2g = 10\text{ m}\cdot\text{s}^{-2}, calculate the magnitude of the electric field intensity EE in NC1\text{N}\cdot\text{C}^{-1}.

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Answer: 50000

Answer

The magnitude of the electric field intensity is 50000 NC150000\text{ N}\cdot\text{C}^{-1} (or 5.0×104 NC15.0 \times 10^4\text{ N}\cdot\text{C}^{-1}).
In electrostatic equilibrium, the sphere experiences three forces: weight (mgmg) vertically downward, electrostatic force (qEqE) horizontally, and tension (TT) along the thread at 4545^\circ to the vertical. Balancing components gives Tsin45=qET \sin 45^\circ = qE and Tcos45=mgT \cos 45^\circ = mg. Dividing these yields tan45=qEmg=1\tan 45^\circ = \frac{qE}{mg} = 1, which gives qE=mgqE = mg. Substituting the given values gives E=2.0×1034.0×108=50000 NC1E = \frac{2.0 \times 10^{-3}}{4.0 \times 10^{-8}} = 50000\text{ N}\cdot\text{C}^{-1}.

Step-by-Step Solution

1
Calculate the weight of the charged sphere.
W=mg=(2.0×104 kg)(10 ms2)=2.0×103 NW = mg = (2.0 \times 10^{-4}\text{ kg})(10\text{ m}\cdot\text{s}^{-2}) = 2.0 \times 10^{-3}\text{ N}.
The weight provides the downward vertical force in equilibrium.
2
Relate the electrostatic force to the weight using the angle of deflection.
tan(45)=FeW    1=Fe2.0×103 N    Fe=2.0×103 N\tan(45^\circ) = \frac{F_e}{W} \implies 1 = \frac{F_e}{2.0 \times 10^{-3}\text{ N}} \implies F_e = 2.0 \times 10^{-3}\text{ N}.
In electrostatic equilibrium, the ratio of the horizontal force to the vertical force equals the tangent of the angle with the vertical.
3
Calculate the electric field strength EE using Fe=qEF_e = qE.
E=Feq=2.0×103 N4.0×108 C=50000 NC1E = \frac{F_e}{q} = \frac{2.0 \times 10^{-3}\text{ N}}{4.0 \times 10^{-8}\text{ C}} = 50000\text{ N}\cdot\text{C}^{-1}.
The electric field intensity is the electric force per unit charge.

Key Concept

Equilibrium of a charged body in a uniform electric field
Estimated Time:2m 0s
Question 7575Question

In ΔPQR\Delta PQR, side p=6 cmp = 6\text{ cm}, side q=62 cmq = 6\sqrt{2}\text{ cm}, and P=30\angle P = 30^\circ. If Q\angle Q is an acute angle, what is the measure of Q\angle Q?

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Answer: 4545^\circ

Answer

4545^\circ
Using the Sine Rule sinQq=sinPp\frac{\sin Q}{q} = \frac{\sin P}{p}, we substitute the given values to find sinQ=62sin306=22\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \frac{\sqrt{2}}{2}. Since Q\angle Q is specified as an acute angle, Q=45\angle Q = 45^\circ.

Step-by-Step Solution

1
Set up the Sine Rule formula relating sides p,qp, q and their opposite angles P,QP, Q.
psinP=qsinQ\frac{p}{\sin P} = \frac{q}{\sin Q}
The Sine Rule allows finding an unknown angle when two side lengths and one opposite angle are known.
2
Substitute the given values p=6p = 6, q=62q = 6\sqrt{2}, and P=30P = 30^\circ into the formula.
6sin30=62sinQ\frac{6}{\sin 30^\circ} = \frac{6\sqrt{2}}{\sin Q}
Populating known quantities permits solving for sinQ\sin Q.
3
Solve for sinQ\sin Q.
sinQ=62sin306=212=22\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}
Simplifying the algebraic fraction yields the exact sine ratio for angle QQ.
4
Determine the acute angle QQ corresponding to sinQ=22\sin Q = \frac{\sqrt{2}}{2}.
Q=45\angle Q = 45^\circ
The principal acute angle with sine equal to 22\frac{\sqrt{2}}{2} is 4545^\circ.

Key Concept

Applying the Sine Rule to calculate an unknown acute angle in a non-right-angled triangle
Question 7576Question

What is the value of the definite integral 13(4x1)dx\int_{1}^{3} (4x - 1) \, dx?

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Answer: 14

Answer

14
The antiderivative of 4x14x - 1 is 2x2x2x^2 - x. Evaluating at the upper boundary x=3x = 3 gives 2(3)23=152(3)^2 - 3 = 15, and at the lower boundary x=1x = 1 gives 2(1)21=12(1)^2 - 1 = 1. Subtracting the lower limit evaluation from the upper limit evaluation yields 151=1415 - 1 = 14.

Step-by-Step Solution

1
Find the indefinite integral of the integrand 4x14x - 1.
(4x1)dx=2x2x\int (4x - 1) \, dx = 2x^2 - x
Using the power rule of integration, xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Substitute the upper limit x=3x = 3 into the antiderivative.
2(3)^2 - (3) = 2(9) - 3 = 18 - 3 = 15
Evaluating F(b)F(b) for F(x)=2x2xF(x) = 2x^2 - x at b=3b = 3.
3
Substitute the lower limit x=1x = 1 into the antiderivative.
2(1)^2 - (1) = 2 - 1 = 1
Evaluating F(a)F(a) for F(x)=2x2xF(x) = 2x^2 - x at a=1a = 1.
4
Subtract the lower limit result from the upper limit result.
15 - 1 = 14
By the Fundamental Theorem of Calculus, abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) \, dx = F(b) - F(a).

Key Concept

Evaluation of Definite Integrals using the Fundamental Theorem of Calculus
Estimated Time:1m 0s
Question 7577Question

A trader bought a set of solar panels for 160,000\text{₦}160,000. He marked up the cost price by 25%25\% to fix the marked price. During a trade fair, he offered a 10%10\% discount on the marked price. What is his net profit in Naira?

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Answer: 20000

Answer

The net profit made by the trader is ���20,000.
First, find the marked price by adding a 25% markup to the cost price: ₦160,000 × 1.25 = ₦200,000. Next, calculate the selling price after a 10% discount on the marked price: ₦200,000 × 0.90 = ₦180,000. Finally, subtract the cost price from the selling price to find the net profit: ₦180,000 - ₦160,000 = ₦20,000.

Step-by-Step Solution

1
Calculate the marked price
Marked Price = ₦200,000
The marked price is 125% of the original cost price of ₦160,000.
2
Calculate the selling price after discount
Selling Price = ₦180,000
A 10% discount reduces the marked price by ₦20,000.
3
Determine the net profit
Net Profit = ₦20,000
Net profit is the difference between the selling price (₦180,000) and the cost price (₦160,000).

Key Concept

Markup, Discount, and Profit Calculations
Question 7578Question

A binary operation \oplus defined on the set of real numbers is given by ab=a+b+kaba \oplus b = a + b + kab, where kk is a non-zero real constant. If the inverse of 33 under this operation is 12-\frac{1}{2}, what is the inverse of 44 under the same operation?

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Answer: 1223-\frac{12}{23}

Answer

1223-\frac{12}{23}
To find the inverse of 44, first determine the identity element ee from ae=aa \oplus e = a, which gives e=0e = 0. Next, set xx1=0x \oplus x^{-1} = 0 to get x1=x1+kxx^{-1} = -\frac{x}{1 + kx}. Using 31=123^{-1} = -\frac{1}{2}, solve 31+3k=12-\frac{3}{1 + 3k} = -\frac{1}{2} to find k=53k = \frac{5}{3}. Finally, substituting x=4x = 4 yields 41=41+4(53)=12234^{-1} = -\frac{4}{1 + 4(\frac{5}{3})} = -\frac{12}{23}.

Step-by-Step Solution

1
Determine the identity element ee of the operation.
e=0e = 0
By definition of identity element, ae=a    a+e+kae=a    e(1+ka)=0    e=0a \oplus e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0 for all valid aa.
2
Derive the general formula for the inverse x1x^{-1} of an element xx.
x1=x1+kxx^{-1} = -\frac{x}{1 + kx}
An element and its inverse yield the identity element under the operation: xx1=0    x+x1+kxx1=0    x1(1+kx)=xx \oplus x^{-1} = 0 \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x.
3
Use the given inverse of 33 to solve for the constant kk.
k=53k = \frac{5}{3}
Given 31=123^{-1} = -\frac{1}{2}, substitute x=3x = 3 into the inverse formula: 31+3k=12    1+3k=6    3k=5    k=53-\frac{3}{1 + 3k} = -\frac{1}{2} \implies 1 + 3k = 6 \implies 3k = 5 \implies k = \frac{5}{3}.
4
Calculate the inverse of 44 using k=53k = \frac{5}{3}.
41=12234^{-1} = -\frac{12}{23}
Substitute x=4x = 4 and k=53k = \frac{5}{3} into the inverse formula: 41=41+4(53)=41+203=4233=12234^{-1} = -\frac{4}{1 + 4\left(\frac{5}{3}\right)} = -\frac{4}{1 + \frac{20}{3}} = -\frac{4}{\frac{23}{3}} = -\frac{12}{23}.

Key Concept

Binary Operations: Identity and Inverse Elements with Unknown Parameters
Estimated Time:2m 0s
Question 7579Question

A pie chart illustrates the annual allocation of funds for an agricultural research station across four sectors: Crop Research, Livestock, Irrigation, and Equipment Maintenance. The sector angles for Crop Research, Livestock, and Irrigation are 135135^\circ, 9090^\circ, and 6060^\circ respectively. If the total annual budget is N7,200,000\text{N}7,200,000, what is the amount allocated to Equipment Maintenance in Naira?

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Answer: 1500000

Answer

The amount allocated to Equipment Maintenance is 1,500,000 Naira.
A complete pie chart has a total angle of 360360^\circ, representing the full amount of N7,200,000\text{N}7,200,000. The angle corresponding to Equipment Maintenance is 360(135+90+60)=75360^\circ - (135^\circ + 90^\circ + 60^\circ) = 75^\circ. The dollar/naira allocation is given by 75360×7,200,000=1,500,000\frac{75^\circ}{360^\circ} \times 7,200,000 = 1,500,000.

Step-by-Step Solution

1
Calculate the sum of the given sector angles
Sum of known angles = 285 degrees
Determining the total angular measure already accounted for by the three known sectors.
2
Subtract the sum of known angles from 360 degrees
Sector angle for Equipment Maintenance = 75 degrees
The sum of all sector angles in a pie chart is always 360 degrees.
3
Multiply the fraction of the pie chart by the total funds
1,500,000 Naira
Converting the sector angle representation to its corresponding quantitative value.

Key Concept

Pie chart sector angle computation and value conversion
Question 7580Question

In a hydrogen atom modeled according to Bohr's theory, an electron undergoes a transition from an excited state with an energy of 1.51 eV-1.51\text{ eV} to a lower energy state of 3.40 eV-3.40\text{ eV}. Calculate the energy of the emitted photon in electron-volts (eV\text{eV}).

Show answer & explanation

Answer: 1.89

Answer

The energy of the emitted photon is 1.89 eV1.89\text{ eV}.
According to Bohr's atomic model, when an electron drops from an initial higher energy level EiE_i to a final lower energy level EfE_f, a photon is emitted carrying energy E=EiEfE = E_i - E_f. Substituting the given values gives E=1.51 eV(3.40 eV)=1.89 eVE = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}.

Step-by-Step Solution

1
Identify the initial and final energy states of the electron.
Ei=1.51 eVE_i = -1.51\text{ eV} and Ef=3.40 eVE_f = -3.40\text{ eV}.
The electron moves from a higher (less negative) energy state to a lower (more negative) energy state.
2
Apply Bohr's energy quantization formula for photon emission Ephoton=EiEfE_{\text{photon}} = E_i - E_f.
Ephoton=1.51(3.40)=1.89 eVE_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}.
By energy conservation, the energy lost by the transitioning electron equals the energy of the emitted photon.

Key Concept

Bohr's Energy Transition Postulate
Estimated Time:1m 0s
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