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1526 soru

Soru 161Soru →

Three of the interior angles of a convex polygon are each 120∘120^\circ, while the remaining interior angles are each 160∘160^\circ. How many sides does the polygon have?

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Cevap: 12

Cevap

The polygon has 12 sides.
The sum of the interior angles of an nn-sided polygon is (n−2)×180∘(n - 2) \times 180^\circ. From the problem, the sum of the angles is 3×120∘+(n−3)×160∘=160∘n−120∘3 \times 120^\circ + (n - 3) \times 160^\circ = 160^\circ n - 120^\circ. Equating 180∘(n−2)180^\circ(n - 2) to 160∘n−120∘160^\circ n - 120^\circ gives 180∘n−360∘=160∘n−120∘180^\circ n - 360^\circ = 160^\circ n - 120^\circ, which simplifies to 20∘n=240∘20^\circ n = 240^\circ, yielding n=12n = 12.

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1
Write down the standard formula for the sum of interior angles of an nn-sided convex polygon.
Sum of interior angles = (n−2)×180∘(n - 2) \times 180^\circ.
Any nn-sided convex polygon can be split into (n−2)(n-2) triangles, each having an interior angle sum of 180∘180^\circ.
2
Express the total sum of interior angles using the given values.
Sum = 3(120∘)+(n−3)(160∘)=360∘+160∘n−480∘=160∘n−120∘3(120^\circ) + (n - 3)(160^\circ) = 360^\circ + 160^\circ n - 480^\circ = 160^\circ n - 120^\circ.
Three angles are 120∘120^\circ, so the remaining (n−3)(n - 3) angles must each equal 160∘160^\circ.
3
Equate the theoretical sum to the calculated sum.
180∘(n−2)=160∘n−120∘  ⟹  180∘n−360∘=160∘n−120∘180^\circ(n - 2) = 160^\circ n - 120^\circ \implies 180^\circ n - 360^\circ = 160^\circ n - 120^\circ.
Both expressions represent the total interior angle sum of the same polygon.
4
Solve the equation for the number of sides nn.
20∘n=240∘  ⟹  n=1220^\circ n = 240^\circ \implies n = 12.
Subtracting 160∘n160^\circ n from both sides and adding 360∘360^\circ yields 20∘n=240∘20^\circ n = 240^\circ.

Anahtar Kavram

Sum of Interior Angles of a Polygon
Soru 162Soru →

The 3rd3^{\text{rd}}, 6th6^{\text{th}}, and 11th11^{\text{th}} terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is 1515, what is the 4th4^{\text{th}} term of the geometric progression?

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Cevap: 125

Cevap

The 4th term of the geometric progression is 125.
By writing the 3rd, 6th, and 11th terms of the AP as 15+2d15+2d, 15+5d15+5d, and 15+10d15+10d, we utilize the geometric mean property (15+5d)2=(15+2d)(15+10d)(15+5d)^2 = (15+2d)(15+10d) to find d=6d=6. This yields the GP terms 27,45,7527, 45, 75, giving a common ratio of 5/35/3. Multiplying the third term 7575 by 5/35/3 gives the 4th GP term as 125125.

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1
Write down the AP term expressions
T3=15+2dT_3 = 15 + 2d, T6=15+5dT_6 = 15 + 5d, T11=15+10dT_{11} = 15 + 10d
The nthn^{\text{th}} term of an AP is given by Tn=a+(n−1)dT_n = a + (n-1)d with initial term a=15a = 15.
2
Apply the geometric progression condition
(15+5d)2=(15+2d)(15+10d)(15 + 5d)^2 = (15 + 2d)(15 + 10d)
If three terms A,B,CA, B, C are in GP, then B2=A⋅CB^2 = A \cdot C.
3
Expand and solve the quadratic equation for the common difference dd
d=6d = 6
Expanding gives 225+150d+25d2=225+180d+20d2  ⟹  5d2=30d  ⟹  d=6225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6 because d≠0d \neq 0.
4
Find the terms and common ratio of the GP
G1=27G_1 = 27, G2=45G_2 = 45, G3=75G_3 = 75, and common ratio r=53r = \frac{5}{3}
Substituting d=6d = 6 gives the GP terms, and dividing consecutive terms gives r=4527=53r = \frac{45}{27} = \frac{5}{3}.
5
Calculate the 4th term of the GP
G4=125G_4 = 125
Multiplying the 3rd term by the common ratio yields 75×53=12575 \times \frac{5}{3} = 125.

Anahtar Kavram

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties
Tahmini Süre:2m 30s
Soru 163Soru →

The perpendicular bisector of the line segment joining the points P(2,−1)P(2, -1) and Q(6,7)Q(6, 7) intersects the yy-axis at (0,c)(0, c). Find the value of cc.

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Cevap: 5

Cevap

The value of cc is 5.
The midpoint of PQPQ is (4,3)(4, 3) and the gradient of PQPQ is 22. The perpendicular bisector has a gradient of −12-\frac{1}{2} and passes through (4,3)(4, 3). Substituting these into the line equation gives y−3=−12(x−4)y - 3 = -\frac{1}{2}(x - 4), which simplifies to y=−12x+5y = -\frac{1}{2}x + 5. The line intersects the yy-axis at (0,5)(0, 5), so c=5c = 5.

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1
Find the midpoint of the line segment PQPQ
Midpoint M=(4,3)M = (4, 3)
The perpendicular bisector must pass through the midpoint of the segment.
2
Calculate the gradient of PQPQ
Gradient mPQ=2m_{PQ} = 2
Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} for points (2,−1)(2, -1) and (6,7)(6, 7).
3
Determine the gradient of the perpendicular line
Perpendicular gradient m⊥=−12m_{\perp} = -\frac{1}{2}
Perpendicular lines have gradients that are negative reciprocals (m1m2=−1m_1 m_2 = -1).
4
Formulate the equation of the perpendicular bisector and solve for the yy-intercept
y=−12x+5y = -\frac{1}{2}x + 5, hence c=5c = 5
Using point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (4,3)(4, 3) and m=−12m = -\frac{1}{2}, setting x=0x = 0 gives the yy-intercept.

Anahtar Kavram

Perpendicular Bisector and Line Equations
Soru 164Soru →

A body of mass 0.2 kg0.2\text{ kg} undergoes simple harmonic motion with an angular frequency of 10 rad/s10\text{ rad/s} and an amplitude of 0.04 m0.04\text{ m}. What is the magnitude of the maximum restoring force acting on the body in newtons?

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Cevap: 0.8

Cevap

The magnitude of the maximum restoring force acting on the body is 0.8 N0.8\text{ N}.
The maximum restoring force in simple harmonic motion occurs at maximum displacement (y=Ay = A) and is given by Fmax=mω2AF_{\text{max}} = m \omega^2 A. Substituting m=0.2 kgm = 0.2\text{ kg}, ω=10 rad/s\omega = 10\text{ rad/s}, and A=0.04 mA = 0.04\text{ m} yields Fmax=0.2×(10)2×0.04=0.8 NF_{\text{max}} = 0.2 \times (10)^2 \times 0.04 = 0.8\text{ N}.

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1
Identify the given physical quantities from the problem statement.
m=0.2 kgm = 0.2\text{ kg}, ω=10 rad/s\omega = 10\text{ rad/s}, and A=0.04 mA = 0.04\text{ m}.
These parameters are required to calculate acceleration and restoring force in simple harmonic motion.
2
Calculate the maximum acceleration of the oscillating body.
amax=ω2A=(10)2×0.04=4.0 m/s2a_{\text{max}} = \omega^2 A = (10)^2 \times 0.04 = 4.0\text{ m/s}^2.
In simple harmonic motion, maximum acceleration occurs at maximum displacement (the amplitude).
3
Determine the maximum restoring force.
Fmax=mamax=0.2×4.0=0.8 NF_{\text{max}} = m a_{\text{max}} = 0.2 \times 4.0 = 0.8\text{ N}.
According to Newton's second law, force is the product of mass and acceleration.

Anahtar Kavram

Maximum Restoring Force in Simple Harmonic Motion
Soru 165Soru →

If 75+12−27=k3\sqrt{75} + \sqrt{12} - \sqrt{27} = k\sqrt{3}, what is the value of kk?

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Cevap: 4

Cevap

The value of kk is 4.
Simplifying each surd into its basic form yields 75=53\sqrt{75} = 5\sqrt{3}, 12=23\sqrt{12} = 2\sqrt{3}, and 27=33\sqrt{27} = 3\sqrt{3}. Combining the coefficients gives (5+2−3)3=43(5 + 2 - 3)\sqrt{3} = 4\sqrt{3}. Equating 434\sqrt{3} to k3k\sqrt{3} shows that k=4k = 4.

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1
Simplify each individual surd by factoring out perfect squares
75=53\sqrt{75} = 5\sqrt{3}, 12=23\sqrt{12} = 2\sqrt{3}, 27=33\sqrt{27} = 3\sqrt{3}
To combine surds through addition or subtraction, they must be converted to similar surds.
2
Combine the coefficients of the like surds
53+23−33=(5+2−3)3=435\sqrt{3} + 2\sqrt{3} - 3\sqrt{3} = (5 + 2 - 3)\sqrt{3} = 4\sqrt{3}
Like terms with the same radical factor 3\sqrt{3} can be added and subtracted directly.
3
Compare the resulting coefficient with k3k\sqrt{3}
k=4k = 4
By direct comparison of coefficients of 3\sqrt{3}, kk equals 4.

Anahtar Kavram

Simplification and combining of similar surds
Soru 166Soru →

Two 2×22 \times 2 matrices are given as A=(2−143)A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and B=(102k)B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}. If the determinant of the product matrix ABAB is equal to 3030, what is the value of kk?

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Cevap: 3

Cevap

The value of k is 3.
The determinant of a product of square matrices equals the product of their individual determinants: det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B). Evaluating det⁡(A)\det(A) yields (2)(3)−(−1)(4)=10(2)(3) - (-1)(4) = 10, and det⁡(B)\det(B) yields (1)(k)−(0)(2)=k(1)(k) - (0)(2) = k. Substituting these into det⁡(AB)=30\det(AB) = 30 gives 10k=3010k = 30, leading directly to k=3k = 3.

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1
Evaluate the determinant of matrix A
\det(A) = 10
Using the 2×22 \times 2 determinant formula det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, we calculate det⁡(A)=(2)(3)−(−1)(4)=6+4=10\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10.
2
Evaluate the determinant of matrix B
\det(B) = k
Calculating the determinant of matrix BB yields det⁡(B)=(1)(k)−(0)(2)=k\det(B) = (1)(k) - (0)(2) = k.
3
Use the product property of determinants to solve for k
k = 3
Since det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B), we have 10k=3010k = 30. Dividing both sides by 10 yields k=3k = 3.

Anahtar Kavram

Determinant of Matrix Product
Tahmini Süre:1m 30s
Soru 167Soru →

The table below shows the frequency distribution of marks obtained by a group of students in a mathematics test:

Mark (xx)246810
Frequency (ff)21412

Find the mean deviation of the distribution.

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Cevap: 2

Cevap

The mean deviation of the distribution is 2.
To find the mean deviation, first calculate the mean xˉ=∑fx∑f=6010=6\bar{x} = \frac{\sum fx}{\sum f} = \frac{60}{10} = 6. Next, sum the absolute deviations multiplied by their frequencies: ∑f∣x−xˉ∣=2(4)+1(2)+4(0)+1(2)+2(4)=20\sum f|x - \bar{x}| = 2(4) + 1(2) + 4(0) + 1(2) + 2(4) = 20. Dividing this total by the sum of frequencies 1010 yields a mean deviation of 22.

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1
Calculate the arithmetic mean of the distribution
\bar{x} = \frac{\sum f x}{\sum f} = \frac{(2 \times 2) + (1 \times 4) + (4 \times 6) + (1 \times 8) + (2 \times 10)}{2 + 1 + 4 + 1 + 2} = \frac{60}{10} = 6
The mean is required as the central benchmark from which individual deviations are measured.
2
Calculate the sum of absolute deviations weighted by frequency
\sum f |x - \bar{x}| = 2|2 - 6| + 1|4 - 6| + 4|6 - 6| + 1|8 - 6| + 2|10 - 6| = 8 + 2 + 0 + 2 + 8 = 20
Each absolute difference from the mean must be multiplied by its frequency to account for the total deviation.
3
Divide the total absolute deviation by the total frequency
\text{Mean Deviation} = \frac{\sum f |x - \bar{x}|}{\sum f} = \frac{20}{10} = 2
The mean deviation represents the average distance of all observations from the arithmetic mean.

Anahtar Kavram

Mean Deviation of a Frequency Distribution
Tahmini Süre:1m 30s
Soru 168Soru →

In the dimensional equation for the period of oscillation of a simple pendulum, T=kgalbT = k g^a l^b, where TT is the period, gg is the acceleration due to gravity, ll is the length of the pendulum, and kk is a dimensionless constant, what is the numerical value of the exponent aa?

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Cevap: -0.5

Cevap

The numerical value of the exponent aa is -0.5.
Applying dimensional analysis to T=kgalbT = k g^a l^b, the dimension of the left-hand side is T1\text{T}^1. The right-hand side has dimensions (L T−2)a(L)b=La+bT−2a(\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}. Equating the exponents of time T\text{T} yields 1=−2a1 = -2a, which gives a=−0.5a = -0.5.

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1
Express the dimensions of all physical quantities involved in fundamental base dimensions (M, L, T).
The dimension of period TT is [T][\text{T}], length ll is [L][\text{L}], and gravitational acceleration gg is [L T−2][\text{L T}^{-2}].
Dimensional analysis requires substituting each quantity with its fundamental dimensions.
2
Formulate the dimensional homogeneity equation.
M0L0T1=(L T−2)a(L)b=La+bT−2a\text{M}^0 \text{L}^0 \text{T}^1 = (\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}.
The principle of dimensional homogeneity states that the exponents of base dimensions on both sides of a physically correct equation must be equal.
3
Equate exponents of time T\text{T} and solve for aa.
1=−2a  ⟹  a=−12=−0.51 = -2a \implies a = -\frac{1}{2} = -0.5.
Comparing the powers of T\text{T} gives a linear equation in aa.

Anahtar Kavram

Principle of Dimensional Homogeneity
Tahmini Süre:1m 15s
Soru 169Soru →

A sector of a circle of radius 14 cm14\text{ cm} subtends an angle of 90∘90^\circ at the centre of the circle. What is the area of the sector in cm2\text{cm}^2? (Take π=227\pi = \frac{22}{7})

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Cevap: 154

Cevap

The area of the sector is 154 cm2154\text{ cm}^2.
The area of a circular sector is given by Area=θ360∘×πr2\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2. Substituting θ=90∘\theta = 90^\circ, r=14 cmr = 14\text{ cm}, and π=227\pi = \frac{22}{7} yields 90360×227×142=14×616=154 cm2\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2.

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1
Identify the formula for the area of a circular sector.
Area=θ360∘×πr2\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
The area of a sector is proportional to the central angle it subtends relative to a full circle (360∘360^\circ).
2
Substitute the given values into the formula.
Area=90∘360∘×227×(14)2\text{Area} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2
Given radius r=14 cmr = 14\text{ cm}, angle θ=90∘\theta = 90^\circ, and π=227\pi = \frac{22}{7}.
3
Simplify the fraction and calculate the numerical value.
Area=14×227×196=14×22×28=22×7=154 cm2\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{1}{4} \times 22 \times 28 = 22 \times 7 = 154\text{ cm}^2
Simplifying 90360\frac{90}{360} yields 14\frac{1}{4} and dividing 196196 by 77 gives 2828.

Anahtar Kavram

Area of a sector of a circle
Tahmini Süre:45s
Soru 170Soru →

A curve is defined by the equation y=x2−6x+11y = x^2 - 6x + 11. What is the minimum value of yy on this curve?

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Cevap: 2

Cevap

The minimum value of yy on the curve is 2.
Differentiating y=x2−6x+11y = x^2 - 6x + 11 gives dydx=2x−6\frac{dy}{dx} = 2x - 6. Setting this derivative to zero yields 2x−6=02x - 6 = 0, so x=3x = 3. Substituting x=3x = 3 into the original function gives y=(3)2−6(3)+11=2y = (3)^2 - 6(3) + 11 = 2. Since d2ydx2=2>0\frac{d^2y}{dx^2} = 2 > 0, the point at x=3x = 3 is a local minimum, making 2 the minimum value of yy.

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1
Find the first derivative of the curve function.
dydx=2x−6\frac{dy}{dx} = 2x - 6
Stationary points occur where the derivative is equal to zero.
2
Solve for the xx-coordinate at the stationary point.
2x - 6 = 0 \implies x = 3
Setting the derivative to zero determines the input value where the slope is horizontal.
3
Calculate the corresponding yy-value at x=3x = 3.
y = (3)^2 - 6(3) + 11 = 2
Evaluating the original equation at x=3x = 3 yields the minimum value of yy.

Anahtar Kavram

Finding the minimum value of a function using differentiation
Soru 171Soru →

In an experiment to determine the density of a solid sphere, the mass is measured as (50.0±0.5) g(50.0 \pm 0.5)\text{ g} and the radius is measured as (2.00±0.05) cm(2.00 \pm 0.05)\text{ cm}. What is the percentage error in the calculated density of the sphere?

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Cevap: 8.5

Cevap

The percentage error in the calculated density of the sphere is 8.5%8.5\%.
The density formula ρ=3m4πr3\rho = \frac{3m}{4\pi r^3} dictates that maximum relative error is given by Δρρ=Δmm+3(Δrr)\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right). Evaluating the percentage errors gives 1.0%1.0\% for mass and 2.5%2.5\% for radius. Summing 1.0%+3(2.5%)1.0\% + 3(2.5\%) yields 8.5%8.5\%.

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1
Determine the functional dependence of density on measured quantities.
Density ρ=mV=3m4πr3\rho = \frac{m}{V} = \frac{3m}{4\pi r^3}.
The volume of a sphere of radius rr is V=43πr3V = \frac{4}{3}\pi r^3.
2
Formulate the maximum fractional error relationship.
\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right).
When combining uncertainties, fractional errors add, and exponents act as multiplying factors.
3
Calculate the percentage error in the mass measurement.
0.550.0×100%=1.0%.\frac{0.5}{50.0} \times 100\% = 1.0\%.
Percentage error in mass is the absolute uncertainty divided by the measured value multiplied by 100%100\%.
4
Calculate the percentage error in the radius measurement.
0.052.00×100%=2.5%.\frac{0.05}{2.00} \times 100\% = 2.5\%.
Percentage error in radius is the absolute uncertainty divided by the measured value multiplied by 100%100\%.
5
Calculate total percentage error in density.
Percentage error = 1.0\% + 3(2.5\%) = 8.5\%.
The radius contributes three times its relative error because volume depends on r3r^3.

Anahtar Kavram

Error propagation in derived quantities involving powers
Soru 172Soru →

A point PP lies inside a circle of radius 13 cm13\text{ cm} at a distance of 5 cm5\text{ cm} from the center OO. A chord ABAB passes through point PP such that the ratio of segment APAP to segment PBPB is 1:41:4. Calculate the total length of the chord ABAB in centimeters.

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Cevap: 30

Cevap

The total length of chord ABAB is 30 cm30\text{ cm}.
Using the Power of a Point property for an interior point PP, the product of the chord segments is AP⋅PB=R2−OP2=132−52=144AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 144. Given AP:PB=1:4AP : PB = 1 : 4, we write AP=xAP = x and PB=4xPB = 4x, leading to 4x2=144  ⟹  x=6 cm4x^2 = 144 \implies x = 6\text{ cm}. Summing the two segments gives AB=6+24=30 cmAB = 6 + 24 = 30\text{ cm}.

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1
Calculate the constant product of chord segments passing through interior point PP.
AP⋅PB=R2−OP2=132−52=169−25=144AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144.
By the intersecting chords theorem, the product of segments created by an interior point PP on any chord equals (R−d)(R+d)=R2−d2(R - d)(R + d) = R^2 - d^2.
2
Set up an algebraic equation using the segment ratio AP:PB=1:4AP : PB = 1 : 4.
Let AP=xAP = x and PB=4xPB = 4x, giving (x)(4x)=144  ⟹  4x2=144(x)(4x) = 144 \implies 4x^2 = 144.
Expressing both chord segments in terms of a single variable xx allows direct calculation of the segment lengths.
3
Solve for xx to find the individual segment lengths.
x2=36  ⟹  x=6 cmx^2 = 36 \implies x = 6\text{ cm}. Therefore, AP=6 cmAP = 6\text{ cm} and PB=24 cmPB = 24\text{ cm}.
Taking the positive square root gives the scale factor xx since physical distances must be positive.
4
Sum the segment lengths to find the total chord length.
AB=AP+PB=6+24=30 cmAB = AP + PB = 6 + 24 = 30\text{ cm}.
The entire chord length is the sum of its two divided parts.

Anahtar Kavram

Intersecting Chords Theorem and Power of an Interior Point
Soru 173Soru →

Given that the determinant of the 3×33 \times 3 matrix M=(3102x−1042)M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix} is equal to 2020, calculate the value of xx.

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Cevap: 2

Cevap

The value of xx is 22.
Expanding the matrix MM along its top row yields det⁡(M)=3(2x+4)−1(4)=6x+8\det(M) = 3(2x + 4) - 1(4) = 6x + 8. Setting this expression equal to 2020 gives 6x+8=206x + 8 = 20, which simplifies to 6x=126x = 12, yielding x=2x = 2.

Adım Adım Çözüm

1
Expand the 3x3 matrix along the first row
\det(M) = 3(2x - (-4)) - 1(4 - 0) + 0
Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.
2
Simplify the algebraic expression for the determinant
\det(M) = 6x + 8
Distribute the coefficients and combine like constant terms.
3
Solve the linear equation for x
x = 2
Subtract 8 from 20 to get 12, then divide by 6.

Anahtar Kavram

Determinant of a 3x3 Matrix via Cofactor Expansion
Soru 174Soru →

A solid object of volume 0.002 m30.002\text{ m}^3 is completely immersed in water of density 1000 kg/m31000\text{ kg/m}^3. What is the magnitude of the upthrust exerted on the object by the water? [Take g=10 m/s2g = 10\text{ m/s}^2]

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Cevap: 20

Cevap

The magnitude of the upthrust exerted on the object is 20 N20\text{ N}.
According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using U=V⋅ρ⋅gU = V \cdot \rho \cdot g. Substituting V=0.002 m3V = 0.002\text{ m}^3, ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3, and g=10 m/s2g = 10\text{ m/s}^2 yields U=0.002×1000×10=20 NU = 0.002 \times 1000 \times 10 = 20\text{ N}.

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1
Identify the given quantities from the problem statement.
V=0.002 m3V = 0.002\text{ m}^3, ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3, and g=10 m/s2g = 10\text{ m/s}^2.
These are the essential inputs required to calculate the weight of the displaced liquid.
2
Apply Archimedes' principle to find upthrust force.
U=Vρg=0.002×1000×10=20 NU = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}.
Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Anahtar Kavram

Archimedes' Principle and Upthrust
Soru 175Soru →

What is the numerical value of the simplified expression 45−1−5\frac{4}{\sqrt{5} - 1} - \sqrt{5}?

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Cevap: 1

Cevap

The numerical value of the expression is 1.
Multiplying the top and bottom of 45−1\frac{4}{\sqrt{5} - 1} by its conjugate (5+1)(\sqrt{5} + 1) simplifies the fraction to 4(5+1)4=5+1\frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1. Subtracting 5\sqrt{5} from 5+1\sqrt{5} + 1 leaves 1.

Adım Adım Çözüm

1
Rationalize the denominator of the fractional term
5+1\sqrt{5} + 1
Multiply both numerator and denominator by the conjugate (5+1)(\sqrt{5} + 1) to apply the difference of two squares identity (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2 in the denominator.
2
Subtract the remaining surd term
1
Subtract 5\sqrt{5} from 5+1\sqrt{5} + 1, leaving the integer 1.

Anahtar Kavram

Rationalization of Binomial Denominators
Soru 176Soru →

Calculate the area, in square units, of the triangle formed by the straight line 3x+4y−24=03x + 4y - 24 = 0 and the coordinate axes.

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Cevap: 24

Cevap

The area of the triangle formed by the line and the coordinate axes is 24 square units.
To find the area of the triangle bounded by a straight line and the coordinate axes, determine the magnitude of the xx-intercept and yy-intercept. Setting y=0y = 0 in 3x+4y−24=03x + 4y - 24 = 0 gives x=8x = 8. Setting x=0x = 0 gives y=6y = 6. The vertices of the right triangle are at (0,0)(0,0), (8,0)(8,0), and (0,6)(0,6). The area is calculated as 12×8×6=24\frac{1}{2} \times 8 \times 6 = 24 square units.

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1
Find the xx-intercept of the straight line
x=8x = 8, corresponding to the point (8,0)(8, 0)
Setting y=0y = 0 determines where the line crosses the horizontal axis
2
Find the yy-intercept of the straight line
y=6y = 6, corresponding to the point (0,6)(0, 6)
Setting x=0x = 0 determines where the line crosses the vertical axis
3
Compute the area of the right-angled triangle formed with the origin (0,0)(0,0)
Area=12×8×6=24\text{Area} = \frac{1}{2} \times 8 \times 6 = 24
The coordinate axes are perpendicular, making the triangle right-angled with base length 8 and height 6

Anahtar Kavram

Area of a triangle bounded by a straight line and the coordinate axes
Tahmini Süre:1m 0s
Soru 177Soru →

A small business recorded the number of customer inquiries received per day over six consecutive days as follows: 44, 77, 88, 1111, 1313, and 1717. What is the variance of the daily customer inquiries?

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Cevap: 18

Cevap

The variance of the daily customer inquiries is 1818.
To find the variance of the data set {4,7,8,11,13,17}\{4, 7, 8, 11, 13, 17\}, first calculate the mean: xˉ=4+7+8+11+13+176=10\bar{x} = \frac{4+7+8+11+13+17}{6} = 10. Next, compute the squared deviation of each data point from the mean: (4−10)2=36(4-10)^2 = 36, (7−10)2=9(7-10)^2 = 9, (8−10)2=4(8-10)^2 = 4, (11−10)2=1(11-10)^2 = 1, (13−10)2=9(13-10)^2 = 9, and (17−10)2=49(17-10)^2 = 49. Summing these squared deviations gives 108108. Dividing this total by the number of observations (66) yields a variance of 1818.

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1
Calculate the arithmetic mean of the given data set.
xˉ=10\bar{x} = 10
The mean is required as the central point from which deviations are calculated.
2
Determine the squared deviation of each data value from the mean.
(−6)2=36(-6)^2 = 36, (−3)2=9(-3)^2 = 9, (−2)2=4(-2)^2 = 4, 12=11^2 = 1, 32=93^2 = 9, 72=497^2 = 49
Variance measures the average squared distance of data points from the mean.
3
Sum all calculated squared deviations.
∑(x−xˉ)2=36+9+4+1+9+49=108\sum (x - \bar{x})^2 = 36 + 9 + 4 + 1 + 9 + 49 = 108
This provides the total sum of squares for the data set.
4
Divide the total sum of squares by the number of observations (n=6n = 6).
σ2=1086=18\sigma^2 = \frac{108}{6} = 18
Population variance formula is σ2=∑(x−xˉ)2n\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}.

Anahtar Kavram

Variance of Ungrouped Data
Tahmini Süre:1m 30s
Soru 178Soru →

Given that F(x)=∫(3x2−2sin⁡x) dxF(x) = \int (3x^2 - 2\sin x) \, dx and F(0)=6F(0) = 6, what is the value of the constant of integration CC?

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Cevap: 4

Cevap

The value of the constant of integration CC is 44.
Integrating 3x2−2sin⁡x3x^2 - 2\sin x yields F(x)=x3+2cos⁡x+CF(x) = x^3 + 2\cos x + C. Substituting x=0x = 0 gives F(0)=2(1)+C=2+CF(0) = 2(1) + C = 2 + C. Since F(0)=6F(0) = 6, setting 2+C=62 + C = 6 yields C=4C = 4.

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1
Integrate the function f(x)=3x2−2sin⁡xf(x) = 3x^2 - 2\sin x with respect to xx
F(x)=x3+2cos⁡x+CF(x) = x^3 + 2\cos x + C
The antiderivative of 3x23x^2 is x3x^3 and the antiderivative of −2sin⁡x-2\sin x is 2cos⁡x2\cos x.
2
Evaluate F(0)F(0) using the antiderivative expression
F(0)=03+2cos⁡(0)+C=2+CF(0) = 0^3 + 2\cos(0) + C = 2 + C
Since cos⁡(0)=1\cos(0) = 1, the term 2cos⁡(0)2\cos(0) simplifies to 22.
3
Solve for the integration constant CC using F(0)=6F(0) = 6
C=4C = 4
Subtracting 22 from both sides of 2+C=62 + C = 6 yields C=4C = 4.

Anahtar Kavram

Indefinite integration of polynomial and trigonometric functions with initial conditions
Soru 179Soru →

A 0.50 kg0.50\text{ kg} mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is 0.16 J0.16\text{ J} and the force constant of the spring is 32 N/m32\text{ N/m}. What is the speed of the mass, in m/s\text{m/s}, at the instant when the magnitude of its acceleration is 3.84 m/s23.84\text{ m/s}^2?

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Cevap: 0.64

Cevap

The speed of the mass at that instant is 0.64 m/s0.64\text{ m/s}.
Using the relation for total mechanical energy E=12kA2E = \frac{1}{2}kA^2, the amplitude is A=0.10 mA = 0.10\text{ m}. The angular frequency is ω=k/m=8.0 rad/s\omega = \sqrt{k/m} = 8.0\text{ rad/s}. From ∣a∣=ω2∣x∣|a| = \omega^2 |x|, the displacement magnitude when acceleration is 3.84 m/s23.84\text{ m/s}^2 is ∣x∣=0.06 m|x| = 0.06\text{ m}. Substituting these values into v=ωA2−x2v = \omega \sqrt{A^2 - x^2} yields v=8.00.102−0.062=0.64 m/sv = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}.

Adım Adım Çözüm

1
Calculate the angular frequency of the simple harmonic motion
ω=8.0 rad/s\omega = 8.0\text{ rad/s}
The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.
2
Calculate the amplitude of oscillation from total energy
A = 0.10\text{ m}
The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.
3
Find the magnitude of displacement corresponding to the given acceleration
|x| = 0.06\text{ m}
In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.
4
Calculate the speed at this displacement using the SHM velocity-displacement relation
v = 0.64\text{ m/s}
Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Anahtar Kavram

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion
Soru 180Soru →

A projectile is launched from ground level over flat terrain. At time t=2 st = 2\text{ s} after launch, the projectile passes through a point located 60 m60\text{ m} horizontally and 60 m60\text{ m} vertically from its launch point. Taking the acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2 and neglecting air resistance, what is the total horizontal range of the projectile in meters?

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Cevap: 240

Cevap

The total horizontal range of the projectile is 240 m240\text{ m}.
The horizontal motion occurs at a constant velocity of 30 m/s30\text{ m/s} calculated from 60 m2 s\frac{60\text{ m}}{2\text{ s}}. Substituting the vertical position (60 m60\text{ m}) and time (2 s2\text{ s}) into y=uyt−5t2y = u_y t - 5t^2 yields an initial vertical velocity of 40 m/s40\text{ m/s}. The total duration in the air is T=2(40)10=8 sT = \frac{2(40)}{10} = 8\text{ s}. The total horizontal range is therefore 30 m/s×8 s=240 m30\text{ m/s} \times 8\text{ s} = 240\text{ m}.

Adım Adım Çözüm

1
Determine the horizontal component of velocity
vx=30 m/sv_x = 30\text{ m/s}
Horizontal velocity remains constant throughout flight because there is no horizontal acceleration.
2
Determine the initial vertical component of velocity
uy=40 m/su_y = 40\text{ m/s}
Applying the vertical displacement equation y=uyt−12gt2y = u_y t - \frac{1}{2}g t^2 with y=60 my = 60\text{ m}, t=2 st = 2\text{ s}, and g=10 m/s2g = 10\text{ m/s}^2.
3
Calculate the total time of flight
T=8 sT = 8\text{ s}
The projectile completes its full parabolic trajectory when vertical displacement returns to zero, given by T=2uygT = \frac{2 u_y}{g}.
4
Calculate the total horizontal range
R=240 mR = 240\text{ m}
The total range is the product of the constant horizontal velocity component and total time of flight (R=vx×TR = v_x \times T).

Anahtar Kavram

Independence of horizontal and vertical components of projectile motion
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