Tüm alıştırma soruları

13931 soru

Soru 4261Soru

What is the numerical value of the simplified expression 4515\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 451\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 (ab)(a+b)=a2b2(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 4262Soru
The volume flow rate QQ of a viscous liquid flowing through a pipe of radius rr under a pressure gradient ΔPl\frac{\Delta P}{l} is modeled by the equation:
Q=kηxry(ΔPl)zQ = k \eta^x r^y \left(\frac{\Delta P}{l}\right)^z
where η\eta is the coefficient of dynamic viscosity and kk is a dimensionless constant. What are the values of the exponents xx, yy, and zz respectively?
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Cevap: x=1,y=4,z=1x = -1, y = 4, z = 1

Cevap

x=1x = -1, y=4y = 4, and z=1z = 1
The dimensional representation of volume flow rate is [L3T1][L^3 T^{-1}], dynamic viscosity is [ML1T1][M L^{-1} T^{-1}], radius is [L][L], and pressure gradient is [ML2T2][M L^{-2} T^{-2}]. Equating the powers of MM, LL, and TT gives x+z=0x + z = 0, x2z=1-x - 2z = -1, and x+y2z=3-x + y - 2z = 3. Solving these simultaneously yields x=1x = -1, y=4y = 4, and z=1z = 1.

Adım Adım Çözüm

1
Determine the fundamental dimensions of each physical quantity
Volume flow rate Q=VolumeTime=[L3T1]Q = \frac{\text{Volume}}{\text{Time}} = [L^3 T^{-1}];
Dynamic viscosity η=[ML1T1]\eta = [M L^{-1} T^{-1}];
Radius r=[L]r = [L];
Pressure gradient ΔPl=PressureLength=[ML1T2][L]=[ML2T2]\frac{\Delta P}{l} = \frac{\text{Pressure}}{\text{Length}} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}].
Correct base dimensions are required for dimensional analysis.
2
Set up the dimensional equation by substituting the base dimensions into the formula
[L3T1]=[ML1T1]x[L]y[ML2T2]z=Mx+zLx+y2zTx2z[L^3 T^{-1}] = [M L^{-1} T^{-1}]^x [L]^y [M L^{-2} T^{-2}]^z = M^{x+z} L^{-x + y - 2z} T^{-x - 2z}.
The principle of dimensional homogeneity requires both sides of the equation to have matching exponents for MM, LL, and TT.
3
Equate exponents for MM, TT, and LL to form algebraic equations
For MM: x+z=0    z=xx + z = 0 \implies z = -x
For TT: x2z=1-x - 2z = -1
For LL: x+y2z=3-x + y - 2z = 3.
This creates a linear system of equations for the exponents xx, yy, and zz.
4
Solve the system of linear equations
Substituting z=xz = -x into the TT equation: x2(x)=1    x=1-x - 2(-x) = -1 \implies x = -1.
Hence z=(1)=1z = -(-1) = 1.
Substituting x=1x = -1 and z=1z = 1 into the LL equation: (1)+y2(1)=3    1+y2=3    y=4-(-1) + y - 2(1) = 3 \implies 1 + y - 2 = 3 \implies y = 4.
Yields the unique set of exponents x=1,y=4,z=1x = -1, y = 4, z = 1.

Anahtar Kavram

Dimensional Analysis and Homogeneity
Tahmini Süre:2m 0s
Soru 4263Soru

In political science, the mass media is classified as a primary agent of political socialization.

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Cevap: False

Cevap

The statement is False. Mass media is a secondary agent of political socialization, whereas the family is the primary agent.
The statement is false. The family is the fundamental primary agent of political socialization because it imparts foundational civic orientations during early childhood development. Mass media, educational institutions, and civic associations are secondary agents.

Adım Adım Çözüm

1
Recall the distinction between primary and secondary agents of political socialization.
Primary agents (such as the family) influence political values early in childhood through direct, close personal interactions. Secondary agents (such as mass media, schools, peer groups, and political parties) operate later in life or through broader, non-familial channels.
Classification depends on the developmental timing and nature of interaction.
2
Determine the agent category for mass media.
Mass media informs and shapes public opinion among older children and adults, placing it in the secondary agent category.
The mass media reinforces, modifies, or expands existing beliefs formed earlier by primary agents.

Anahtar Kavram

Primary vs. Secondary Agents of Political Socialization
Soru 4264Soru

Calculate the area, in square units, of the triangle formed by the straight line 3x+4y24=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+4y24=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.

Adım Adım Çözüm

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 4265Soru

An open rectangular box with a square base of side length x cmx\text{ cm} is to be constructed such that its total surface area is 108 cm2108\text{ cm}^2. What is the maximum volume of the box in cm3\text{cm}^3?

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Cevap: 108 cm3108\text{ cm}^3

Cevap

108 cm3108\text{ cm}^3
For an open box with a square base of side xx and height hh, total surface area is A=x2+4xh=108 cm2A = x^2 + 4xh = 108\text{ cm}^2. Solving for hh gives h=108x24xh = \frac{108 - x^2}{4x}. Substituting hh into the volume formula gives V(x)=x2h=27x14x3V(x) = x^2 h = 27x - \frac{1}{4}x^3. Differentiating gives dVdx=2734x2\frac{dV}{dx} = 27 - \frac{3}{4}x^2. Setting dVdx=0\frac{dV}{dx} = 0 yields x=6x = 6. The second derivative d2Vdx2=32x\frac{d^2V}{dx^2} = -\frac{3}{2}x evaluated at x=6x = 6 is 9<0-9 < 0, confirming a local maximum. Evaluating V(6)=27(6)14(63)=16254=108 cm3V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3 yields the correct maximum volume.

Adım Adım Çözüm

1
Set up the surface area equation for the open box and express height hh in terms of xx.
Surface area A=x2+4xh=108    h=108x24xA = x^2 + 4xh = 108 \implies h = \frac{108 - x^2}{4x}.
An open box with a square base has 1 base face (x2x^2) and 4 vertical side faces (xhxh).
2
Formulate the volume function V(x)V(x) in terms of xx.
V(x)=x2h=x2(108x24x)=14(108xx3)=27x14x3V(x) = x^2 h = x^2 \left(\frac{108 - x^2}{4x}\right) = \frac{1}{4}(108x - x^3) = 27x - \frac{1}{4}x^3.
Substitute the expression for hh into the volume formula V=x2hV = x^2 h.
3
Find the critical point by differentiating V(x)V(x) with respect to xx and setting dVdx=0\frac{dV}{dx} = 0.
dVdx=2734x2=0    34x2=27    x2=36    x=6 cm\frac{dV}{dx} = 27 - \frac{3}{4}x^2 = 0 \implies \frac{3}{4}x^2 = 27 \implies x^2 = 36 \implies x = 6\text{ cm}.
Stationary points occur where the first derivative of the volume function equals zero.
4
Verify that x=6x = 6 gives a maximum volume and calculate V(6)V(6).
d2Vdx2=32x    d2Vdx2x=6=9<0\frac{d^2V}{dx^2} = -\frac{3}{2}x \implies \left.\frac{d^2V}{dx^2}\right|_{x=6} = -9 < 0 (maximum). Volume V(6)=27(6)14(63)=16254=108 cm3V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3.
The negative second derivative confirms a maximum turning point.

Anahtar Kavram

Optimization of physical quantities using the first and second derivative tests.
Tahmini Süre:2m 0s
Soru 4266Soru

If set A={a,b,c,d,e}A = \{a, b, c, d, e\} and set B={c,d,e,f,g}B = \{c, d, e, f, g\}, what is the number of elements in the set (AB)(BA)(A \setminus B) \cup (B \setminus A)?

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Cevap: 4; four

Cevap

The number of elements in (AB)(BA)(A \setminus B) \cup (B \setminus A) is 4.
The set difference ABA \setminus B consists of elements in AA that are not in BB, which gives {a,b}\{a, b\}. Similarly, BAB \setminus A consists of elements in BB that are not in AA, giving {f,g}\{f, g\}. The union (AB)(BA)(A \setminus B) \cup (B \setminus A) is {a,b,f,g}\{a, b, f, g\}, which has a cardinality of 4.

Adım Adım Çözüm

1
Find the relative difference ABA \setminus B
AB={a,b}A \setminus B = \{a, b\}
Remove elements of BB present in AA.
2
Find the relative difference BAB \setminus A
BA={f,g}B \setminus A = \{f, g\}
Remove elements of AA present in BB.
3
Take the union of the two set differences
(AB)(BA)={a,b,f,g}(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}
Combine elements from both set differences.
4
Count the number of elements in the resulting set
4 elements
The set {a,b,f,g}\{a, b, f, g\} contains 4 distinct elements.

Anahtar Kavram

Symmetric Difference of Two Sets
Tahmini Süre:45s
Soru 4267Soru

A circle is inscribed in an isosceles trapezium ABCDABCD with parallel sides ABAB and CDCD. If AB=18 cmAB = 18\text{ cm} and CD=8 cmCD = 8\text{ cm}, what is the area of the region inside the trapezium that lies outside the circle?

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Cevap: (15636π) cm2(156 - 36\pi)\text{ cm}^2

Cevap

(15636π) cm2(156 - 36\pi)\text{ cm}^2
For an isosceles trapezium with an inscribed circle, the sum of opposite sides must be equal (AB+CD=AD+BC=26 cmAB + CD = AD + BC = 26\text{ cm}), giving slant side length 13 cm13\text{ cm}. Using Pythagoras, the perpendicular distance (height) is 13252=12 cm\sqrt{13^2 - 5^2} = 12\text{ cm}. The area of the trapezium is 12(18+8)(12)=156 cm2\frac{1}{2}(18 + 8)(12) = 156\text{ cm}^2. The inscribed circle has radius equal to half the height (6 cm6\text{ cm}), so its area is π×62=36π cm2\pi \times 6^2 = 36\pi\text{ cm}^2. Subtracting the circle area from the trapezium area yields (15636π) cm2(156 - 36\pi)\text{ cm}^2.

Adım Adım Çözüm

1
Apply the property of a tangential quadrilateral to find the non-parallel sides
For a quadrilateral with an inscribed circle, the sum of opposite sides is equal: AB+CD=AD+BC=18+8=26 cmAB + CD = AD + BC = 18 + 8 = 26\text{ cm}. Since trapezium ABCDABCD is isosceles, AD=BC=13 cmAD = BC = 13\text{ cm}.
Tangential quadrilaterals have equal sums of opposite side lengths.
2
Calculate the height hh of the trapezium using the Pythagorean theorem
Dropping vertical altitudes from top vertices CC and DD creates right triangles at the base with horizontal leg 1882=5 cm\frac{18 - 8}{2} = 5\text{ cm}. Thus, h=13252=16925=12 cmh = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\text{ cm}.
The height of the trapezium forms the perpendicular leg of the right-angled side triangle.
3
Find the area of the trapezium and the inscribed circle
Area of trapezium =12(AB+CD)×h=12(18+8)×12=156 cm2= \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(18 + 8) \times 12 = 156\text{ cm}^2. The diameter of the inscribed circle equals the height h=12 cmh = 12\text{ cm}, so its radius is r=6 cmr = 6\text{ cm}. Area of circle =πr2=36π cm2= \pi r^2 = 36\pi\text{ cm}^2.
The diameter of a circle inscribed between parallel bases equals the vertical height between those bases.
4
Subtract the area of the circle from the area of the trapezium
Remaining Area =15636π cm2= 156 - 36\pi\text{ cm}^2.
The region inside the trapezium but outside the circle is the difference between their areas.

Anahtar Kavram

Perimeter and Area of Composite Figures and Inscribed Shapes
Soru 4268Soru

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: (410)2=36(4-10)^2 = 36, (710)2=9(7-10)^2 = 9, (810)2=4(8-10)^2 = 4, (1110)2=1(11-10)^2 = 1, (1310)2=9(13-10)^2 = 9, and (1710)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.

Adım Adım Çözüm

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.
(xxˉ)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=(xxˉ)2n\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}.

Anahtar Kavram

Variance of Ungrouped Data
Tahmini Süre:1m 30s
Soru 4269Soru

Given that F(x)=(3x22sinx)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 3x22sinx3x^2 - 2\sin x yields F(x)=x3+2cosx+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)=3x22sinxf(x) = 3x^2 - 2\sin x with respect to xx
F(x)=x3+2cosx+CF(x) = x^3 + 2\cos x + C
The antiderivative of 3x23x^2 is x3x^3 and the antiderivative of 2sinx-2\sin x is 2cosx2\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 4270Soru

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=ω2x|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=ωA2x2v = \omega \sqrt{A^2 - x^2} yields v=8.00.1020.062=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 4271Soru

Match each vector scenario on the left with its correct resultant magnitude or value on the right.

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Öğeler

Resultant of two perpendicular forces of magnitudes 6 N6\text{ N} and 8 N8\text{ N}
Minimum possible magnitude of the resultant of two forces of 7 N7\text{ N} and 12 N12\text{ N}
Resultant magnitude of two equal forces of 15 N15\text{ N} inclined at an angle of 120120^\circ to each other
Magnitude of a 3D displacement vector given by r=(3i+4j+12k) m\mathbf{r} = (3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k})\text{ m}

Eşleşmeler

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Cevap

Match perpendicular forces of 6 N and 8 N to 10 N; minimum resultant of 7 N and 12 N forces to 5 N; two 15 N forces at 120 degrees to 15 N; and 3D displacement vector to 13 m.
Each vector calculation correctly applies the geometric or algebraic properties of vectors: orthogonal vector resolution, opposite-direction subtraction, law of cosines for equal magnitudes at 120 degrees, and 3D component magnitude synthesis.

Adım Adım Çözüm

1
Calculate the magnitude of perpendicular vectors
R=62+82=10 NR = \sqrt{6^2 + 8^2} = 10\text{ N}
Perpendicular vectors form a right-angled triangle, so the Pythagorean theorem applies.
2
Find the minimum resultant magnitude of two vectors
Rmin=12 N7 N=5 NR_{\text{min}} = 12\text{ N} - 7\text{ N} = 5\text{ N}
Minimum resultant occurs when vectors act collinear in opposite directions.
3
Determine the resultant of two equal vectors at 120120^\circ
R=15 NR = 15\text{ N}
Using the cosine rule R=F2+F2+2F2cos120R = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ}, since cos120=0.5\cos 120^\circ = -0.5, R=FR = F.
4
Compute the magnitude of the 3D displacement vector
r=32+42+122=13 m|\mathbf{r}| = \sqrt{3^2 + 4^2 + 12^2} = 13\text{ m}
3D magnitude is calculated using the square root of the sum of squared orthogonal components.

Anahtar Kavram

Vector Addition, Resolution, and Magnitude Evaluation
Soru 4272Soru

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=uyt5t2y = 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=uyt12gt2y = 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
Soru 4273Soru

A Vernier caliper with a least count of 0.01 cm0.01\text{ cm} is used to measure the internal diameter of a beaker. Before making the measurement, the jaws are brought together tightly; the zero mark of the Vernier scale lies to the left of the zero mark of the main scale, and the 6th6\text{th} division on the Vernier scale coincides with a main scale division. If the observed reading for the internal diameter of the beaker is 3.43 cm3.43\text{ cm}, what is the actual internal diameter of the beaker?

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Cevap: 3.47 cm3.47\text{ cm}

Cevap

The actual internal diameter of the beaker is 3.47 cm3.47\text{ cm}.
The instrument possesses a negative zero error because the Vernier zero mark lies to the left of the main scale zero mark when closed. The magnitude of this negative error is calculated as (106)×0.01 cm=0.04 cm-(10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}. Applying the standard correction formula Actual Value=Observed ValueZero Error\text{Actual Value} = \text{Observed Value} - \text{Zero Error} gives 3.43 cm(0.04 cm)=3.47 cm3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}.

Adım Adım Çözüm

1
Determine the zero error of the Vernier caliper
Zero error = (106)×0.01 cm=0.04 cm- (10 - 6) \times 0.01\text{ cm} = -0.04\text{ cm}
When the zero mark of the Vernier scale lies to the left of the main scale zero mark, the zero error is negative. For a 10-division Vernier scale, the magnitude is given by (10N)×least count(10 - N) \times \text{least count}, where N=6N=6 is the coinciding division.
2
Apply the zero error correction formula
Actual Reading = Observed Reading - Zero Error
The true reading is obtained by subtracting the zero error (with its sign) from the observed value.
3
Calculate the actual internal diameter
Actual Diameter = 3.43 cm(0.04 cm)=3.47 cm3.43\text{ cm} - (-0.04\text{ cm}) = 3.47\text{ cm}
Subtracting a negative zero error is mathematically equivalent to adding its magnitude to the observed reading.

Anahtar Kavram

Vernier Caliper Zero Error Correction
Tahmini Süre:1m 30s
Soru 4274Soru

A ship uses a sonar device to determine the depth of the ocean floor. A sound pulse sent vertically downward reflects off the seabed and is detected 0.6 s0.6\text{ s} after transmission. If the speed of sound in seawater is 1500 m/s1500\text{ m/s}, what is the depth of the ocean floor at that location?

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Cevap: 450 m450\text{ m}

Cevap

450 m450\text{ m}
The correct answer is 450 m450\text{ m} because sound travels to the ocean floor and back, covering twice the actual depth. Dividing the total distance of 900 m900\text{ m} by 22 yields the one-way depth of 450 m450\text{ m}.

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1
Calculate the total distance traveled by the sound wave
dtotal=1500 m/s×0.6 s=900 md_{\text{total}} = 1500\text{ m/s} \times 0.6\text{ s} = 900\text{ m}
Total distance traveled by the wave equals speed multiplied by total elapsed time.
2
Determine the one-way depth of the ocean floor
Depth=dtotal2=900 m2=450 m\text{Depth} = \frac{d_{\text{total}}}{2} = \frac{900\text{ m}}{2} = 450\text{ m}
An echo travels down to the seabed and back to the receiver, so the ocean depth is half the total distance covered.

Anahtar Kavram

Echo distance calculation
Tahmini Süre:45s
Soru 4275Soru

Five daily rainfall measurements (in mm) recorded in a city are 3,6,7,9,3, 6, 7, 9, and 1515. What is the standard deviation of these rainfall measurements?

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Cevap: 44

Cevap

The standard deviation of the rainfall measurements is 44.
First compute the mean of the data set: xˉ=3+6+7+9+155=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8. Next, sum the squared deviations from the mean: (38)2+(68)2+(78)2+(98)2+(158)2=25+4+1+1+49=80(3-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2 + (15-8)^2 = 25 + 4 + 1 + 1 + 49 = 80. Dividing by the number of observations (55) yields the variance σ2=805=16\sigma^2 = \frac{80}{5} = 16. Taking the square root gives the standard deviation σ=16=4\sigma = \sqrt{16} = 4.

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1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset
xˉ=3+6+7+9+155=405=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = \frac{40}{5} = 8
The mean is needed to find deviations for each data point.
2
Find the deviation of each number from the mean and square it
(38)2=25(3-8)^2 = 25, (68)2=4(6-8)^2 = 4, (78)2=1(7-8)^2 = 1, (98)2=1(9-8)^2 = 1, (158)2=49(15-8)^2 = 49
Squaring ensures all deviation values are non-negative.
3
Sum the squared deviations and divide by the total number of items (n=5n = 5) to find the variance
\text{Variance } (\sigma^2) = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16
Variance measures the average squared distance from the mean.
4
Take the square root of the variance to obtain standard deviation
\text{Standard Deviation } (\sigma) = \sqrt{16} = 4
Standard deviation returns the dispersion measure back to the original units.

Anahtar Kavram

Standard Deviation of Ungrouped Data
Soru 4276Soru

A train accelerates uniformly along a straight track from an initial velocity of 10 m/s10\text{ m/s} to a final velocity of 30 m/s30\text{ m/s} over a distance of 100 m100\text{ m}. What is the magnitude of the acceleration of the train?

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Cevap: 4 m/s24\text{ m/s}^2

Cevap

4 m/s24\text{ m/s}^2
Using the equation of motion v2=u2+2asv^2 = u^2 + 2as, substitute u=10 m/su = 10\text{ m/s}, v=30 m/sv = 30\text{ m/s}, and s=100 ms = 100\text{ m}. This gives 302=102+2(a)(100)30^2 = 10^2 + 2(a)(100), which simplifies to 900100=200a900 - 100 = 200a, or 800=200a800 = 200a. Solving for acceleration yields a=4 m/s2a = 4\text{ m/s}^2.

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1
Identify the given kinematic values from the problem statement.
Initial velocity u=10 m/su = 10\text{ m/s}, final velocity v=30 m/sv = 30\text{ m/s}, and displacement s=100 ms = 100\text{ m}.
Listing known values helps in selecting the appropriate equation of motion.
2
Select the linear motion formula relating uu, vv, ss, and acceleration aa.
v2=u2+2asv^2 = u^2 + 2as
This formula connects initial velocity, final velocity, distance, and acceleration without requiring time tt.
3
Substitute the values into the equation and solve for aa.
302=102+2(a)(100)    900=100+200a    800=200a    a=4 m/s230^2 = 10^2 + 2(a)(100) \implies 900 = 100 + 200a \implies 800 = 200a \implies a = 4\text{ m/s}^2
Algebraic rearrangement yields the magnitude of acceleration.

Anahtar Kavram

Equations of Uniformly Accelerated Motion
Tahmini Süre:45s
Soru 4277Soru

A water pump driven by an engine with an efficiency of 80%80\% raises water from an underground tank of depth 20 m20\text{ m} and discharges it through a nozzle of cross-sectional area 10 cm210\text{ cm}^2 at a steady speed of 10 m/s10\text{ m/s}. Taking the density of water as 1000 kg/m31000\text{ kg/m}^3 and acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the minimum input power rating (in W\text{W}) required for the engine?

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Cevap: 3125

Cevap

The minimum input power rating required for the engine is 3125 W3125\text{ W}.
The engine must supply power to lift 10 kg10\text{ kg} of water per second through a vertical height of 20 m20\text{ m} while accelerating it to 10 m/s10\text{ m/s}. The useful power output is 2000 W2000\text{ W} (potential) +500 W+ 500\text{ W} (kinetic) =2500 W= 2500\text{ W}. Accounting for an engine efficiency of 80%80\%, the total input power is 25000.80=3125 W\frac{2500}{0.80} = 3125\text{ W}.

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1
Determine the mass of water discharged per unit time (mass flow rate).
dmdt=ρ×A×v=1000 kg/m3×(10×104 m2)×10 m/s=10 kg/s\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}
Water is moving through a cross-sectional area at a constant velocity.
2
Calculate the useful output power required to lift the water and impart kinetic energy.
Pout=dmdtgh+12dmdtv2=(10×10×20)+(12×10×102)=2000 W+500 W=2500 WP_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}
The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.
3
Calculate the total input power using engine efficiency.
Pin=PoutEfficiency=2500 W0.80=3125 WP_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}
Efficiency is the ratio of useful power output to total power input.

Anahtar Kavram

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations
Soru 4278Soru

What is the remainder when the polynomial P(x)=x3+3x22x+4P(x) = x^3 + 3x^2 - 2x + 4 is divided by x1x - 1?

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Cevap: 66

Cevap

The remainder when P(x)P(x) is divided by x1x - 1 is 66.
According to the Remainder Theorem, dividing a polynomial P(x)P(x) by a linear divisor xax - a leaves a remainder equal to P(a)P(a). For the divisor x1x - 1, setting x1=0x - 1 = 0 yields x=1x = 1. Substituting x=1x = 1 into P(x)=x3+3x22x+4P(x) = x^3 + 3x^2 - 2x + 4 gives 1+32+4=61 + 3 - 2 + 4 = 6. Therefore, the value 66 is the correct remainder.

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1
Apply the Remainder Theorem
To find the remainder when P(x)P(x) is divided by xax - a, set x1=0x - 1 = 0, giving x=1x = 1. The remainder is equal to P(1)P(1).
By the Remainder Theorem, dividing a polynomial P(x)P(x) by (xa)(x - a) yields a remainder of P(a)P(a).
2
Substitute x=1x = 1 into P(x)=x3+3x22x+4P(x) = x^3 + 3x^2 - 2x + 4
P(1)=(1)3+3(1)22(1)+4=1+32+4=6P(1) = (1)^3 + 3(1)^2 - 2(1) + 4 = 1 + 3 - 2 + 4 = 6.
Direct evaluation of the expression at x=1x = 1 yields the numerical value of the remainder.

Anahtar Kavram

The Remainder Theorem states that when a polynomial P(x)P(x) is divided by a linear factor (xa)(x - a), the remainder is P(a)P(a).
Soru 4279Soru

Examine the poetic extract below and complete the literary analysis statement by identifying the correct figure of speech.

Aşağıdaki boşlukları doldurun

"The western wave was all a-flame,
The day was well nigh done!"
— Samuel Taylor Coleridge, The Rime of the Ancient Mariner

In the excerpt above, the word 'wave' is employed to represent the entire ocean. The figure of speech in which a part of an entity is used to stand for the whole is known as
.
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Cevap

synecdoche
Synecdoche is a figure of speech where a part of something represents the whole. In Coleridge's verse, 'wave' (an individual component of the body of water) is used to signify the entire ocean.

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1
Analyze the target word 'wave' in relation to the surrounding context of the poem.
The word 'wave' is used by the poet to denote the entire sea or ocean.
A wave is a constituent physical part of the sea.
2
Classify the specific trope that uses a part-for-whole substitution.
Synecdoche is defined as substituting a part for the whole (or the whole for a part).
Because 'wave' (a part) represents the ocean (the whole), the literary device is synecdoche.

Anahtar Kavram

Synecdoche vs. Metonymy in Poetic Imagery
Tahmini Süre:2m 0s
Soru 4280Soru

Match each physical wave description on the left with its definitive wave classification on the right based on propagation mechanisms and oscillation directions.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Oscillations of mutually perpendicular electric and magnetic fields requiring no material medium for propagation
Periodic compressions and rarefactions of medium particles vibrating parallel to the direction of wave travel
Superposition of two identical progressive waves moving in opposite directions, resulting in localized energy without net propagation
Distortions in an elastic medium where particles vibrate perpendicularly to the direction of outward energy propagation

Eşleşmeler

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Cevap

The correct pairings are: Electromagnetic field oscillations without a medium match with Non-mechanical transverse progressive wave; Periodic compressions/rarefactions parallel to wave motion match with Longitudinal mechanical progressive wave; Superposition of opposing identical waves with zero net energy flow matches with Stationary (standing) mechanical wave; Perpendicular particle vibrations in an elastic medium match with Transverse mechanical progressive wave.
Each wave type is categorized by three primary criteria: medium requirement (mechanical requires a medium, non-mechanical does not), vibration alignment (transverse is perpendicular, longitudinal is parallel), and energy movement (progressive transfers energy continuously, stationary confines energy between nodes). Oscillating fields in vacuum represent non-mechanical transverse waves; compressions/rarefactions represent longitudinal mechanical waves; opposing wave superposition forms stationary waves; and perpendicular elastic vibrations represent transverse mechanical waves.

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1
Classify the wave mechanism based on medium requirement.
Waves requiring an elastic medium (particle vibrations) are mechanical; waves capable of propagating through a vacuum via field oscillations are non-mechanical.
Determines whether the wave is mechanical or non-mechanical.
2
Classify the direction of particle or field oscillation relative to propagation direction.
Oscillations parallel to wave motion form longitudinal waves; oscillations perpendicular to wave motion form transverse waves.
Distinguishes between longitudinal and transverse wave modes.
3
Determine energy transfer characteristics (progressive vs. stationary).
Continuous outward transfer of energy indicates a progressive wave, whereas trapped energy bounded by fixed nodes/antinodes resulting from wave superposition indicates a standing (stationary) wave.
Differentiates traveling waves from standing wave patterns.
4
Map each wave description to its complete wave classification.
All four wave descriptions are uniquely matched with their corresponding taxonomy classifications.
Completes the matching process.

Anahtar Kavram

Classification of waves by medium requirement, oscillation direction, and energy propagation mode
Tahmini Süre:1m 30s
ÖncekiSayfa 214 / 697Sonraki
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