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Question 7681Question →

A straight line L1L_1 passes through the points (k,2)(k, 2) and (4,8)(4, 8). If L1L_1 is perpendicular to the line 2x−3y+6=02x - 3y + 6 = 0, what is the value of kk?

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Answer: 8

Answer

The value of kk is 88.
The line 2x−3y+6=02x - 3y + 6 = 0 has a gradient of 23\frac{2}{3}. A line perpendicular to it must have a gradient of −32-\frac{3}{2}. Calculating the gradient of L1L_1 using the points (k,2)(k, 2) and (4,8)(4, 8) gives 8−24−k=64−k\frac{8-2}{4-k} = \frac{6}{4-k}. Equating 64−k=−32\frac{6}{4-k} = -\frac{3}{2} and solving for kk gives k=8k = 8.

Step-by-Step Solution

1
Find the gradient of the given line 2x−3y+6=02x - 3y + 6 = 0
Rearranging into y=mx+cy = mx + c form gives 3y=2x+6  ⟹  y=23x+23y = 2x + 6 \implies y = \frac{2}{3}x + 2. Therefore, the gradient m2=23m_2 = \frac{2}{3}.
To find the perpendicular gradient, we first need the gradient of the given line.
2
Determine the gradient of line L1L_1
Since L1L_1 is perpendicular to the given line, its gradient m1m_1 satisfies m1⋅m2=−1m_1 \cdot m_2 = -1. Thus, m1=−12/3=−32m_1 = -\frac{1}{2/3} = -\frac{3}{2}.
Perpendicular lines have gradients that are negative reciprocals of each other.
3
Express the gradient of L1L_1 using the coordinates (k,2)(k, 2) and (4,8)(4, 8) and solve for kk
Gradient formula: m1=y2−y1x2−x1=8−24−k=64−km_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{4 - k} = \frac{6}{4 - k}. Setting this equal to −32-\frac{3}{2}:
64−k=−32\frac{6}{4 - k} = -\frac{3}{2}
6⋅2=−3(4−k)6 \cdot 2 = -3(4 - k)
12=−12+3k12 = -12 + 3k
3k=24  ⟹  k=83k = 24 \implies k = 8.
Equating the slope calculated from points to the perpendicular slope allows solving for the unknown coordinate kk.

Key Concept

Perpendicular gradients and line slope formula
Question 7682Question →

A progressive wave traveling along a medium is described by the equation y=0.05sin⁡(20πt−4πx)y = 0.05 \sin(20\pi t - 4\pi x), where xx and yy are measured in meters and tt is in seconds. What is the speed of the wave in m/s\text{m/s}?

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Answer: 5

Answer

The speed of the wave is 5.0 m/s5.0\text{ m/s}.
Comparing y=0.05sin⁡(20πt−4πx)y = 0.05 \sin(20\pi t - 4\pi x) to the standard wave equation y=Asin⁡(ωt−kx)y = A \sin(\omega t - kx), we find ω=20π rad/s\omega = 20\pi\text{ rad/s} and k=4π rad/mk = 4\pi\text{ rad/m}. The wave speed vv is calculated as v=ωk=20π4π=5.0 m/sv = \frac{\omega}{k} = \frac{20\pi}{4\pi} = 5.0\text{ m/s}.

Step-by-Step Solution

1
Identify the wave parameters from the standard equation form
ω=20π rad/s\omega = 20\pi\text{ rad/s} and k=4π rad/mk = 4\pi\text{ rad/m}
Matching the given equation y=0.05sin⁡(20πt−4πx)y = 0.05 \sin(20\pi t - 4\pi x) to y=Asin⁡(ωt−kx)y = A \sin(\omega t - kx) gives the values for angular frequency ω\omega and wave number kk.
2
Compute wave speed using the relationship v=ωkv = \frac{\omega}{k}
v=20π4π=5.0 m/sv = \frac{20\pi}{4\pi} = 5.0\text{ m/s}
Wave speed is defined as the ratio of angular frequency to wave number.

Key Concept

Wave Speed from Wave Equation
Question 7683Question →

Five of the interior angles of a convex polygon are each equal to 140∘140^\circ, while the remaining interior angles are each equal to 160∘160^\circ. Calculate the number of sides of the polygon.

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Answer: 13

Answer

The number of sides of the polygon is 13.
Each 140∘140^\circ interior angle has an exterior angle of 40∘40^\circ, contributing 5×40∘=200∘5 \times 40^\circ = 200^\circ to the exterior angle sum. Each 160∘160^\circ interior angle has an exterior angle of 20∘20^\circ, contributing (n−5)×20∘(n - 5) \times 20^\circ. Since the sum of exterior angles of any convex polygon is 360∘360^\circ, setting 200+20(n−5)=360200 + 20(n - 5) = 360 yields 20n=26020n = 260, giving n=13n = 13.

Step-by-Step Solution

1
Calculate the exterior angle measures
The exterior angles are 180∘−140∘=40∘180^\circ - 140^\circ = 40^\circ (for 5 vertices) and 180∘−160∘=20∘180^\circ - 160^\circ = 20^\circ (for the remaining n−5n - 5 vertices).
Interior and exterior angles at each vertex of a polygon form a linear pair and sum to 180∘180^\circ.
2
Apply the sum of exterior angles property
5(40∘)+(n−5)(20∘)=360∘5(40^\circ) + (n - 5)(20^\circ) = 360^\circ.
The sum of exterior angles of any convex polygon is always constant and equal to 360∘360^\circ.
3
Solve the linear equation for nn
200+20n−100=360  ⟹  20n=260  ⟹  n=13200 + 20n - 100 = 360 \implies 20n = 260 \implies n = 13.
Expanding terms and isolating nn gives the exact number of sides.

Key Concept

Exterior angle sum property of convex polygons
Question 7684Question →

A circle of radius 7 cm7\text{ cm} is inscribed inside a square. What is the area of the region inside the square but outside the circle? (Take π=227\pi = \frac{22}{7})

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Answer: 42 cm242\text{ cm}^2

Answer

The area of the region inside the square but outside the circle is 42 cm242\text{ cm}^2.
Because the circle is inscribed inside the square, the diameter of the circle is equal to the side length of the square: s=2r=2(7)=14 cms = 2r = 2(7) = 14\text{ cm}. The area of the square is 142=196 cm214^2 = 196\text{ cm}^2. The area of the circle is πr2=227×72=154 cm2\pi r^2 = \frac{22}{7} \times 7^2 = 154\text{ cm}^2. Subtracting the area of the circle from the area of the square gives the region inside the square but outside the circle: 196−154=42 cm2196 - 154 = 42\text{ cm}^2.

Step-by-Step Solution

1
Determine the side length of the square.
Side length s=2×7 cm=14 cms = 2 \times 7\text{ cm} = 14\text{ cm}.
An inscribed circle touches all four sides of the square, so its diameter equals the side length of the square.
2
Calculate the area of the square.
Area of square =s2=142=196 cm2= s^2 = 14^2 = 196\text{ cm}^2.
The area of a square with side length ss is given by s2s^2.
3
Calculate the area of the inscribed circle.
Area of circle =πr2=227×72=22×7=154 cm2= \pi r^2 = \frac{22}{7} \times 7^2 = 22 \times 7 = 154\text{ cm}^2.
The area of a circle with radius rr is given by πr2\pi r^2.
4
Subtract the circle's area from the square's area to find the area of the shaded outer region.
Area =196 cm2−154 cm2=42 cm2= 196\text{ cm}^2 - 154\text{ cm}^2 = 42\text{ cm}^2.
The remaining area consists of the four corner regions between the circle and the bounding square.

Key Concept

Area of composite plane figures (shaded area between inscribed shape and container)
Question 7685Question →

Find the positive value of xx that satisfies the simultaneous equations y−2x=3y - 2x = 3 and y=x2+3y = x^2 + 3.

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Answer: 22

Answer

The positive value of xx is 2.
Rearranging the linear equation gives y=2x+3y = 2x + 3. Setting this equal to the second expression for yy gives 2x+3=x2+32x + 3 = x^2 + 3, which simplifies to x2−2x=0x^2 - 2x = 0. Factoring out xx yields x(x−2)=0x(x - 2) = 0. The solutions for xx are 00 and 22. The positive solution is 22.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation
y=2x+3y = 2x + 3
Isolating yy prepares for substitution into the quadratic equation.
2
Substitute y=2x+3y = 2x + 3 into the second equation y=x2+3y = x^2 + 3
2x+3=x2+32x + 3 = x^2 + 3
This creates a single quadratic equation in terms of xx.
3
Rearrange and factor the quadratic equation
x2−2x=0  ⟹  x(x−2)=0x^2 - 2x = 0 \implies x(x - 2) = 0
Subtracting 33 and 2x2x from both sides simplifies the equation to factorable form.
4
Solve for the non-zero (positive) value of xx
x=2x = 2
Setting x−2=0x - 2 = 0 yields x=2x = 2.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Question 7686Question →

Two identical positive point charges, each of magnitude q=+2.5×10−6 Cq = +2.5 \times 10^{-6}\text{ C}, are fixed in a vacuum at a distance of 0.60 m0.60\text{ m} apart. A third point charge q0=+1.0×10−6 Cq_0 = +1.0 \times 10^{-6}\text{ C} is placed on the perpendicular bisector of the line joining the two fixed charges, at a distance of 0.40 m0.40\text{ m} from their midpoint. Taking Coulomb's constant k=9.0×109 N m2 C−2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}, what is the magnitude of the net electrostatic force acting on the third charge, in newtons?

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Answer: 0.144

Answer

The magnitude of the net electrostatic force acting on the third charge is 0.144 N0.144\text{ N}.
Each fixed charge exerts an equal repulsive electrostatic force of 0.09 N0.09\text{ N} on the third charge. Due to the symmetrical arrangement, the force components perpendicular to the bisector cancel each other out, while the parallel components add together, giving a net force of 2×0.09×0.8=0.144 N2 \times 0.09 \times 0.8 = 0.144\text{ N}.

Step-by-Step Solution

1
Determine the distance from each fixed charge to the third charge
r=0.50 mr = 0.50\text{ m}
The charges form a right-angled triangle with base 0.30 m0.30\text{ m} (half of 0.60 m0.60\text{ m}) and height 0.40 m0.40\text{ m}, yielding a hypotenuse of 0.302+0.402=0.50 m\sqrt{0.30^2 + 0.40^2} = 0.50\text{ m}.
2
Calculate the magnitude of the individual repulsive force from one charge
F=0.09 NF = 0.09\text{ N}
Applying Coulomb's law: F=kqq0r2=9.0×109×2.5×10−6×1.0×10−60.25=0.09 NF = \frac{k q q_0}{r^2} = \frac{9.0 \times 10^9 \times 2.5 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.25} = 0.09\text{ N}.
3
Determine directional component of forces along the perpendicular bisector
cos⁡θ=0.8\cos\theta = 0.8
The directional cosine along the axis of symmetry is the ratio of the adjacent side (0.40 m0.40\text{ m}) to the hypotenuse (0.50 m0.50\text{ m}).
4
Compute the net electrostatic force using vector addition
Fnet=0.144 NF_{\text{net}} = 0.144\text{ N}
Horizontal components cancel by symmetry, so Fnet=2Fcos⁡θ=2×0.09×0.8=0.144 NF_{\text{net}} = 2 F \cos\theta = 2 \times 0.09 \times 0.8 = 0.144\text{ N}.

Key Concept

Vector superposition of Coulombic forces along an axis of symmetry
Question 7687Question →

A student library committee is selecting 55 distinct books from a shelf containing 77 novel titles and 55 biography titles. If 22 specific novel titles are mutually exclusive (they cannot both be selected together in the same combination), in how many ways can the selection of 55 books be made such that at least 33 novel titles are included?

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Answer: 436

Answer

436
To find the number of valid ways, we first find the total number of ways to select at least 33 novels from 77 novels and 55 biographies, which equals 546546. Next, we determine how many of these combinations contain both of the restricted novels while still having at least 33 novels overall. There are 110110 such invalid selections. Subtracting 110110 from 546546 gives 436436 valid ways.

Step-by-Step Solution

1
Calculate the total combinations with at least 3 novels without any restriction.
Case 1 (33 Novels, 22 Biographies): (73)×(52)=35×10=350\binom{7}{3} \times \binom{5}{2} = 35 \times 10 = 350.
Case 2 (44 Novels, 11 Biography): (74)×(51)=35×5=175\binom{7}{4} \times \binom{5}{1} = 35 \times 5 = 175.
Case 3 (55 Novels, 00 Biographies): (75)×(50)=21×1=21\binom{7}{5} \times \binom{5}{0} = 21 \times 1 = 21.
Total unrestricted combinations = 350+175+21=546350 + 175 + 21 = 546.
Establishing the total pool of choices that satisfy the constraint of selecting at least 3 novels.
2
Calculate the invalid combinations where both restricted novels are selected together AND at least 3 novels are included.
If both restricted novels are selected, we have already chosen 22 novels. To reach a total of 55 books with at least 33 novels, we must pick 33 additional books from the remaining 1010 books (55 remaining novels and 55 biographies), EXCLUDING the scenario where all 33 additional books are biographies (which would leave us with only 22 novels in total).
Combinations of 33 books from 1010 remaining books = (103)=120\binom{10}{3} = 120.
Combinations of 33 biographies from 55 biographies = (53)=10\binom{5}{3} = 10.
Invalid combinations = 120−10=110120 - 10 = 110.
To apply the mutual exclusion constraint correctly, we must subtract only those invalid selections that also satisfy the condition of having at least 3 novels.
3
Subtract invalid combinations from total unrestricted combinations.
Valid combinations = 546−110=436546 - 110 = 436.
Subtracting the forbidden overlapping outcomes yields the net valid choices.

Key Concept

Combinations with multiple constraints and mutual exclusion
Question 7688Question →

A plane mechanical wave propagating through a fluid is represented by the equation y=0.02sin⁡(250πt−5π2x)y = 0.02 \sin \left(250\pi t - \frac{5\pi}{2} x\right), where xx and yy are in meters and tt is in seconds. When the wave passes into a secondary fluid medium, its speed decreases to 60 m/s60\text{ m/s}. Assuming the frequency remains constant, what is the wavelength of the wave in the secondary medium?

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Answer: 0.48 m0.48\text{ m}

Answer

The wavelength of the wave in the secondary medium is 0.48 m0.48\text{ m}.
Comparing the given equation to the general progressive wave equation y=Asin⁡(ωt−kx)y = A \sin(\omega t - kx), we find the angular frequency ω=250π rad/s\omega = 250\pi\text{ rad/s}. The wave frequency is f=ω2π=125 Hzf = \frac{\omega}{2\pi} = 125\text{ Hz}. Because wave frequency is invariant across media boundaries, ff remains 125 Hz125\text{ Hz} in the secondary medium. Using the wave equation v=fλv = f \lambda, the wavelength in the secondary fluid is λ=vf=60 m/s125 Hz=0.48 m\lambda = \frac{v}{f} = \frac{60\text{ m/s}}{125\text{ Hz}} = 0.48\text{ m}.

Step-by-Step Solution

1
Extract angular frequency and wave number from the progressive wave equation
From y=0.02sin⁡(250πt−5π2x)y = 0.02 \sin \left(250\pi t - \frac{5\pi}{2} x\right), ω=250π rad/s\omega = 250\pi\text{ rad/s} and k=5π2 rad/mk = \frac{5\pi}{2}\text{ rad/m}.
Standard progressive wave equations follow the format y=Asin⁡(ωt−kx)y = A \sin(\omega t - kx).
2
Determine the frequency of the wave
f=ω2π=250π2π=125 Hzf = \frac{\omega}{2\pi} = \frac{250\pi}{2\pi} = 125\text{ Hz}.
Frequency is related to angular frequency by ω=2πf\omega = 2\pi f.
3
Apply boundary transition rules to determine the new wavelength
Frequency ff remains constant at 125 Hz125\text{ Hz} across media. Therefore, λ2=v2f=60 m/s125 Hz=0.48 m\lambda_2 = \frac{v_2}{f} = \frac{60\text{ m/s}}{125\text{ Hz}} = 0.48\text{ m}.
When a wave travels across different media boundaries, its frequency depends only on the source and remains constant, whereas speed and wavelength adjust accordingly.

Key Concept

Wave equation parameter extraction and frequency invariance during refraction
Estimated Time:1m 30s
Question 7689Question →

A resistance thermometer registers a resistance of 5.0 Ω5.0\,\Omega at the ice point (0∘C0^\circ\text{C}) and 25.0 Ω25.0\,\Omega at the steam point (100∘C100^\circ\text{C}). When placed in a heated liquid bath, the resistance measured is 30.0 Ω30.0\,\Omega. What is the temperature of the bath on the Kelvin scale?

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Answer: 398 K398\,\text{K}

Answer

The temperature of the bath on the Kelvin scale is 398 K398\,\text{K}.
Using the linear interpolation formula for a thermometric property θ=Rθ−R0R100−R0×100∘C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}, substituting R0=5.0 ΩR_0 = 5.0\,\Omega, R100=25.0 ΩR_{100} = 25.0\,\Omega, and Rθ=30.0 ΩR_\theta = 30.0\,\Omega yields θ=25.020.0×100=125∘C\theta = \frac{25.0}{20.0} \times 100 = 125^\circ\text{C}. Converting to absolute thermodynamic temperature gives T=125+273=398 KT = 125 + 273 = 398\,\text{K}.

Step-by-Step Solution

1
Calculate the temperature on the Celsius scale using linear interpolation of thermometric property.
θ=Rθ−R0R100−R0×100∘C=30.0−5.025.0−5.0×100∘C=25.020.0×100∘C=125∘C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{30.0 - 5.0}{25.0 - 5.0} \times 100^\circ\text{C} = \frac{25.0}{20.0} \times 100^\circ\text{C} = 125^\circ\text{C}
The change in resistance is directly proportional to the temperature change between fixed points.
2
Convert the temperature from degrees Celsius to Kelvins.
T=θ+273=125+273=398 KT = \theta + 273 = 125 + 273 = 398\,\text{K}
Absolute temperature in Kelvin is obtained by adding 273 to the temperature in degrees Celsius.

Key Concept

Temperature Scale Interpolation and Kelvin Conversion
Estimated Time:1m 30s
Question 7690Question →

A music festival coordinator needs to select 44 bands to perform from a pool of 88 available bands. If 22 specific bands insist on either both being selected or neither being selected, in how many different ways can the 44 bands be chosen?

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Answer: 3030

Answer

The total number of ways to choose the bands under the given condition is 3030.
To satisfy the condition that the two specific bands are either both chosen or neither chosen, we evaluate two distinct cases. Case 1 (both chosen) requires choosing 22 additional bands from the remaining 66, yielding 6C2=15{}^6C_2 = 15 ways. Case 2 (neither chosen) requires choosing all 44 bands from the remaining 66, yielding 6C4=15{}^6C_4 = 15 ways. Adding both cases gives 15+15=3015 + 15 = 30 ways.

Step-by-Step Solution

1
Analyze Case 1: Both specific bands are selected.
If both specific bands are included, we only need to select 22 more bands from the remaining 66 bands. The number of ways is 6C2=6×52×1=15{}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15.
Since order does not matter in forming a group of performers, we use combinations.
2
Analyze Case 2: Neither of the specific bands is selected.
If neither of the 22 specific bands is chosen, all 44 bands must be selected from the remaining 66 bands. The number of ways is 6C4=6C2=15{}^6C_4 = {}^6C_2 = 15.
Excluding the 22 specific bands leaves 66 candidate bands to choose 44 from.
3
Sum the mutually exclusive cases.
\text{Total ways} = 15 + 15 = 30.
The two scenarios are disjoint, so the addition principle applies.

Key Concept

Combinations with conditional restrictions
Estimated Time:1m 30s
Question 7691Question →

Find the sum of all integer values of xx that satisfy both the linear inequality 2x−1≥32x - 1 \ge 3 and the quadratic inequality x2−5x−14≤0x^2 - 5x - 14 \le 0.

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Answer: 27

Answer

The sum of all integer values of xx satisfying both inequalities is 2727.
Solving the linear inequality 2x−1≥32x - 1 \ge 3 yields x≥2x \ge 2. Solving the quadratic inequality x2−5x−14≤0x^2 - 5x - 14 \le 0 by factoring gives (x−7)(x+2)≤0(x - 7)(x + 2) \le 0, which defines the interval −2≤x≤7-2 \le x \le 7. Taking the intersection of x≥2x \ge 2 and −2≤x≤7-2 \le x \le 7 gives 2≤x≤72 \le x \le 7. The integer values satisfying this range are 2,3,4,5,6,2, 3, 4, 5, 6, and 77, and their sum is 2727.

Step-by-Step Solution

1
Solve the linear inequality
2x≥4  ⟹  x≥22x \ge 4 \implies x \ge 2
Adding 1 to both sides and dividing by 2 isolates the variable xx.
2
Factor and solve the quadratic inequality
(x−7)(x+2)≤0  ⟹  −2≤x≤7(x - 7)(x + 2) \le 0 \implies -2 \le x \le 7
The roots of the quadratic equation are x=7x = 7 and x=−2x = -2. The parabola opens upward, so the expression is non-positive between the roots.
3
Determine the intersection of both solution sets
2≤x≤72 \le x \le 7
The values of xx must simultaneously satisfy x≥2x \ge 2 and −2≤x≤7-2 \le x \le 7.
4
List all integer solutions within the valid interval
x∈{2,3,4,5,6,7}x \in \{2, 3, 4, 5, 6, 7\}
These are all the whole numbers contained in the closed interval [2,7][2, 7].
5
Sum the integer solutions
2+3+4+5+6+7=272 + 3 + 4 + 5 + 6 + 7 = 27
Summing the identified integer values yields the final required numerical answer.

Key Concept

Linear and Quadratic Inequalities
Estimated Time:1m 30s
Question 7692Question →

A physical quantity ZZ is defined by the expression Z=P⋅VI⋅tZ = \frac{P \cdot V}{I \cdot t}, where PP represents pressure, VV represents volume, II represents electric current, and tt represents time. Which of the following statements correctly classifies quantity ZZ and expresses it in fundamental SI base units?

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Answer: Quantity ZZ is a derived quantity with fundamental SI base units of kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.

Answer

Quantity ZZ is a derived quantity with fundamental SI base units of kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.
The quantity ZZ represents electric potential (voltage), which is defined as work done per unit electric charge (W/QW/Q). Since electric potential is derived from mass, length, time, and electric current, it is a derived physical quantity. Expanding P⋅VP \cdot V yields energy units (kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}), and dividing by I⋅tI \cdot t (A⋅s\text{A}\cdot\text{s}) gives the correct base units of kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.

Step-by-Step Solution

1
Determine the SI base unit decomposition for pressure (PP) and volume (VV).
Pressure P=ForceArea=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2P = \frac{\text{Force}}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}. Volume V=m3V = \text{m}^3. Therefore, P⋅V=(kg⋅m−1⋅s−2)⋅m3=kg⋅m2⋅s−2P \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}.
Decomposing composite physical quantities into base units of mass, length, and time.
2
Determine the SI base unit decomposition for the denominator I⋅tI \cdot t.
I⋅t=Electric current×Time=A⋅sI \cdot t = \text{Electric current} \times \text{Time} = \text{A}\cdot\text{s}.
Electric current (amperes, A) and time (seconds, s) are fundamental SI quantities.
3
Evaluate the full expression for Z=P⋅VI⋅tZ = \frac{P \cdot V}{I \cdot t} in SI base units.
Base units of Z=kg⋅m2⋅s−2A⋅s=kg⋅m2⋅s−3⋅A−1Z = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.
Apply laws of indices to simplify base unit ratios.
4
Classify physical quantity ZZ.
Because ZZ (electric potential) is defined in terms of fundamental quantities (kg, m, s, A) rather than existing independently, it is a derived quantity.
The seven fundamental SI quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Key Concept

Classification of physical quantities into fundamental and derived types and resolving complex derived quantities into base SI units.
Question 7693Question →

A trigonometric function is defined as f(x)=asin⁡(bx)+cf(x) = a \sin(b x) + c, where a>0a > 0 and b>0b > 0. The graph of y=f(x)y = f(x) has a maximum value of 88, a minimum value of −2-2, and a period of 120∘120^\circ. What is the value of a+b+ca + b + c?

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Answer: 11

Answer

The value of a+b+ca + b + c is 11.
Using the maximum (a+c=8a + c = 8) and minimum (−a+c=−2-a + c = -2), solving simultaneously gives a=5a = 5 and c=3c = 3. Using the period formula T=360∘b=120∘T = \frac{360^\circ}{b} = 120^\circ, we obtain b=3b = 3. Adding these values together yields 5+3+3=115 + 3 + 3 = 11.

Step-by-Step Solution

1
Set up a system of linear equations for the amplitude and vertical shift from the maximum and minimum bounds
a=5a = 5 and c=3c = 3
Since sin⁡(bx)\sin(bx) ranges from −1-1 to 11, the maximum is a(1)+c=8a(1) + c = 8 and minimum is a(−1)+c=−2a(-1) + c = -2. Solving these simultaneous equations gives c=3c = 3 and a=5a = 5.
2
Calculate the frequency coefficient bb using the period formula
b=3b = 3
For a sine curve specified in degrees, the period is T=360∘bT = \frac{360^\circ}{b}. Substituting T=120∘T = 120^\circ yields b=3b = 3.
3
Compute the requested sum a+b+ca + b + c
a+b+c=11a + b + c = 11
Summing the calculated parameters 5+3+3=115 + 3 + 3 = 11.

Key Concept

Determining parameters of a trigonometric graph from amplitude, vertical shift, and period.
Question 7694Question →

A parallel plate air capacitor of capacitance 8.0 μF8.0\text{ }\mu\text{F} is fully charged using a 40.0 V40.0\text{ V} d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of 4.04.0 is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules (μJ\mu\text{J}).

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Answer: 4800

Answer

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.
Disconnecting the battery ensures that the charge Q=C0V=320 μCQ = C_0 V = 320\text{ }\mu\text{C} on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to 32.0 μF32.0\text{ }\mu\text{F}. The energy decreases from Ui=6400 μJU_i = 6400\text{ }\mu\text{J} to Uf=Q22Cf=1600 μJU_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}, giving a total decrease of 4800 μJ4800\text{ }\mu\text{J}.

Step-by-Step Solution

1
Calculate the initial energy stored in the air capacitor before disconnection.
Initial energy Ui=12C0V2=12×8.0×10−6 F×(40.0 V)2=6.4×10−3 J=6400 μJU_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}.
The initial state has known capacitance and potential difference.
2
Calculate the new capacitance with the dielectric present.
Final capacitance Cf=KC0=4.0×8.0 μF=32.0 μFC_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}.
Inserting a dielectric of constant KK scales the capacitance by KK.
3
Calculate the final stored energy using charge conservation.
Final energy Uf=UiK=6400 μJ4.0=1600 μJU_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}.
Disconnection forces the charge QQ to remain fixed, so energy scales inversely with capacitance (U=Q22CU = \frac{Q^2}{2C}).
4
Find the difference between initial and final energy.
\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.
The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Key Concept

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions
Estimated Time:2m 0s
Question 7695Question →

In atomic physics, the energy EE of a photon is related to its frequency ff by the formula E=hfE = h f, where hh is Planck's constant. What are the fundamental dimensions of Planck's constant hh?

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Answer: ML2T−1M L^2 T^{-1}

Answer

The fundamental dimensions of Planck's constant are ML2T−1M L^2 T^{-1}.
Planck's constant hh is given by h=E/fh = E / f. Since energy EE has fundamental dimensions of ML2T−2M L^2 T^{-2} and frequency ff has dimensions of T−1T^{-1}, dividing energy by frequency yields ML2T−2T−1=ML2T−1\frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}.

Step-by-Step Solution

1
Express Planck's constant in terms of energy and frequency
h=Efh = \frac{E}{f}
Rearranging the equation E=hfE = h f isolates hh.
2
Determine the fundamental dimensions of energy (EE) and frequency (ff)
[E]=ML2T−2[E] = M L^2 T^{-2} and [f]=T−1[f] = T^{-1}
Energy has dimensions of work (F×d=MLT−2⋅L=ML2T−2F \times d = M L T^{-2} \cdot L = M L^2 T^{-2}) and frequency is inverse time (T−1T^{-1}).
3
Divide the dimensions of energy by the dimensions of frequency
[h]=ML2T−2T−1=ML2T−2−(−1)=ML2T−1[h] = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-2 - (-1)} = M L^2 T^{-1}
Applying the rules of indices simplifies the exponent of time.

Key Concept

Dimensional Analysis of Physical Constants
Estimated Time:1m 0s
Question 7696Question →

An electric charge of +3.2×10−19 C+3.2 \times 10^{-19}\text{ C} is situated in a uniform electric field of strength 5.0×104 N C−15.0 \times 10^4\text{ N C}^{-1}. What is the magnitude of the electrostatic force exerted on the charge?

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Answer: 1.6×10−14 N1.6 \times 10^{-14}\text{ N}

Answer

The magnitude of the electrostatic force exerted on the charge is 1.6×10−14 N1.6 \times 10^{-14}\text{ N}.
The relationship between electric field intensity EE, charge qq, and electrostatic force FF is given by F=qEF = qE. Multiplying +3.2×10−19 C+3.2 \times 10^{-19}\text{ C} by 5.0×104 N C−15.0 \times 10^4\text{ N C}^{-1} gives 1.6×10−14 N1.6 \times 10^{-14}\text{ N}, which correctly expresses the force magnitude in standard notation.

Step-by-Step Solution

1
Identify the given values and formula.
Charge q=3.2×10−19 Cq = 3.2 \times 10^{-19}\text{ C}, Electric field strength E=5.0×104 N C−1E = 5.0 \times 10^4\text{ N C}^{-1}. Formula: F=qEF = qE.
The force experienced by a charge in an electric field is the product of the charge magnitude and the field strength.
2
Substitute the values into the formula and calculate.
F=(3.2×10−19)×(5.0×104)=16.0×10−15 N=1.6×10−14 NF = (3.2 \times 10^{-19}) \times (5.0 \times 10^4) = 16.0 \times 10^{-15}\text{ N} = 1.6 \times 10^{-14}\text{ N}.
Multiplying the numerical coefficients (3.2×5.0=16.0)(3.2 \times 5.0 = 16.0) and combining the powers of ten (10−19×104=10−15)(10^{-19} \times 10^4 = 10^{-15}) gives 1.6×10−14 N1.6 \times 10^{-14}\text{ N} in standard scientific notation.

Key Concept

Electric Field Intensity and Electrostatic Force (F=qEF = qE)
Question 7697Question →

A fixed line segment ABAB has a length of 8 cm8\text{ cm}. A point PP moves in the plane such that the area of △PAB\triangle PAB is always 20 cm220\text{ cm}^2. Which of the following best describes the locus of PP?

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Answer: A pair of parallel lines at a perpendicular distance of 5 cm5\text{ cm} on opposite sides of ABAB

Answer

A pair of parallel lines at a perpendicular distance of 5 cm5\text{ cm} on opposite sides of ABAB
The area of △PAB\triangle PAB is given by 12×base×height\frac{1}{2} \times \text{base} \times \text{height}. With a base AB=8 cmAB = 8\text{ cm}, an area of 20 cm220\text{ cm}^2 requires a constant height h=5 cmh = 5\text{ cm}. The geometric locus of all points at a constant distance from a given straight line consists of two parallel lines situated at that distance on either side of the line.

Step-by-Step Solution

1
Express the area formula of the triangle in terms of base and height.
Area=12×base×h=12×8×h=4h\text{Area} = \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 8 \times h = 4h
The base of △PAB\triangle PAB is fixed as the length of segment ABAB, which is 8 cm8\text{ cm}.
2
Calculate the constant perpendicular height hh.
4h=20  ⟹  h=5 cm4h = 20 \implies h = 5\text{ cm}
Setting the calculated area equal to the given constant area of 20 cm220\text{ cm}^2 gives the required height.
3
Determine the geometric locus corresponding to a constant perpendicular height.
The locus of points at a fixed distance h=5 cmh = 5\text{ cm} from line ABAB is a pair of parallel lines running on either side of ABAB at a distance of 5 cm5\text{ cm}.
Any point PP lying on either of these two parallel lines maintains a perpendicular distance of 5 cm5\text{ cm} from ABAB, ensuring Area(△PAB)=20 cm2\text{Area}(\triangle PAB) = 20\text{ cm}^2.

Key Concept

Locus at a constant distance from a straight line
Question 7698Question →

Find the equation of the normal to the curve y=2sin⁡x−cos⁡xy = 2\sin x - \cos x at the point where x=0x = 0.

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Answer: x+2y+2=0x + 2y + 2 = 0

Answer

x+2y+2=0x + 2y + 2 = 0
At x=0x = 0, the yy-coordinate is 2sin⁡(0)−cos⁡(0)=−12\sin(0) - \cos(0) = -1. Evaluating the derivative dydx=2cos⁡x+sin⁡x\frac{dy}{dx} = 2\cos x + \sin x at x=0x = 0 yields a tangent slope of 22. Since the normal is perpendicular to the tangent, its gradient is −12-\frac{1}{2}. Substituting the point (0,−1)(0, -1) and slope −12-\frac{1}{2} into the line formula yields x+2y+2=0x + 2y + 2 = 0.

Step-by-Step Solution

1
Find the yy-coordinate of the point of contact
At x=0x = 0, y=2sin⁡(0)−cos⁡(0)=0−1=−1y = 2\sin(0) - \cos(0) = 0 - 1 = -1. The point is (0,−1)(0, -1).
The line equation requires a point (x1,y1)(x_1, y_1) on the curve.
2
Differentiate the curve to find dydx\frac{dy}{dx}
dydx=2cos⁡x−(−sin⁡x)=2cos⁡x+sin⁡x\frac{dy}{dx} = 2\cos x - (-\sin x) = 2\cos x + \sin x.
The derivative gives the gradient function of the curve.
3
Calculate the gradient of the tangent and normal at x=0x = 0
Tangent gradient mt=2cos⁡(0)+sin⁡(0)=2(1)+0=2m_t = 2\cos(0) + \sin(0) = 2(1) + 0 = 2. Normal gradient mn=−1mt=−12m_n = -\frac{1}{m_t} = -\frac{1}{2}.
The normal line is perpendicular to the tangent line.
4
Form the equation of the normal line
y−(−1)=−12(x−0)  ⟹  y+1=−12x  ⟹  2y+2=−x  ⟹  x+2y+2=0y - (-1) = -\frac{1}{2}(x - 0) \implies y + 1 = -\frac{1}{2}x \implies 2y + 2 = -x \implies x + 2y + 2 = 0.
Apply the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1).

Key Concept

Equation of Normal to a Curve
Estimated Time:1m 30s
Question 7699Question →

If 132x−54x=45x132_x - 54_x = 45_x, where xx is a positive integer base, what is the value of xx?

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Answer: 7

Answer

The value of the base xx is 7.
Expanding all numbers in terms of powers of xx yields 132x=x2+3x+2132_x = x^2 + 3x + 2, 54x=5x+454_x = 5x + 4, and 45x=4x+545_x = 4x + 5. Substituting these into 132x−54x=45x132_x - 54_x = 45_x gives (x2+3x+2)−(5x+4)=4x+5(x^2 + 3x + 2) - (5x + 4) = 4x + 5. Simplifying this equation results in x2−6x−7=0x^2 - 6x - 7 = 0. Factoring (x−7)(x+1)=0(x - 7)(x + 1) = 0 gives solutions x=7x = 7 and x=−1x = -1. Since a base must be a positive integer greater than 5, the base xx is 7.

Step-by-Step Solution

1
Convert each term from base xx to base 10 using place-value expansion.
132x=1⋅x2+3⋅x1+2⋅x0=x2+3x+2132_x = 1 \cdot x^2 + 3 \cdot x^1 + 2 \cdot x^0 = x^2 + 3x + 2, 54x=5x+454_x = 5x + 4, and 45x=4x+545_x = 4x + 5.
Converting all terms to a common base (base 10) allows standard algebraic solving.
2
Substitute the expanded expressions into the given equation.
(x2+3x+2)−(5x+4)=4x+5(x^2 + 3x + 2) - (5x + 4) = 4x + 5.
Set up the algebraic equation in terms of xx.
3
Simplify the equation into standard quadratic form.
x2−2x−2=4x+5  ⟹  x2−6x−7=0x^2 - 2x - 2 = 4x + 5 \implies x^2 - 6x - 7 = 0.
Combine like terms and move all terms to one side.
4
Solve the quadratic equation for xx.
(x−7)(x+1)=0  ⟹  x=7(x - 7)(x + 1) = 0 \implies x = 7 or x=−1x = -1.
Factor the quadratic expression.
5
Select the valid base.
x=7x = 7.
A number base must be a positive integer greater than the largest digit present in the equation (which is 5).

Key Concept

Solving equations involving unknown number bases by expanding in powers of the base.
Question 7700Question →

The derived SI unit of electrical resistance is the ohm (Ω\Omega). Which of the following correctly expresses the ohm strictly in terms of fundamental SI base units?

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Answer: kg⋅m2⋅s−3⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

Answer

kg⋅m2⋅s−3⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}
Resistance is defined by Ohm's Law as voltage divided by current (R=V/IR = V / I). Voltage is work done per unit charge (V=W/QV = W / Q), and charge is current multiplied by time (Q=I⋅tQ = I \cdot t). Substituting the unit of work (kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}) gives R=kg⋅m2⋅s−2A2⋅s=kg⋅m2⋅s−3⋅A−2R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}.

Step-by-Step Solution

1
Relate resistance to potential difference and current using Ohm's Law.
R=VIR = \frac{V}{I}
Resistance is defined as the ratio of potential difference to electric current.
2
Express electric potential difference (VV) in terms of work (WW) and charge (QQ), where Q=I⋅tQ = I \cdot t.
V=WI⋅t  ⟹  R=WI2⋅tV = \frac{W}{I \cdot t} \implies R = \frac{W}{I^2 \cdot t}
Electric potential difference is work done per unit charge, and electric charge is current multiplied by time.
3
Break down work (W=Force×distanceW = \text{Force} \times \text{distance}) into base SI units.
Unit of Work = kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}
Force is mass times acceleration (kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}), so work is kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}.
4
Substitute all base units into the resistance formula.
Unit of R=kg⋅m2⋅s−2A2⋅s=kg⋅m2⋅s−3⋅A−2R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2 \cdot \text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}
Combining powers of base units yields the complete fundamental SI base unit expression for the ohm.

Key Concept

Derivation of SI base units from defining formulas of physical quantities
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