Tüm alıştırma soruları

13931 soru

Soru 7841Soru

The mean mass of 66 packages in a delivery van is 20 kg20\text{ kg}. If one package weighing 35 kg35\text{ kg} is unloaded from the van, what is the mean mass of the remaining 55 packages?

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Cevap: 17 kg17\text{ kg}

Cevap

The mean mass of the remaining packages is 17 kg17\text{ kg}.
The total weight of the initial 66 packages is 6×20 kg=120 kg6 \times 20\text{ kg} = 120\text{ kg}. Unloading one package of 35 kg35\text{ kg} leaves 12035=85 kg120 - 35 = 85\text{ kg}. Dividing this remaining mass by the remaining 55 packages yields 855=17 kg\frac{85}{5} = 17\text{ kg}.

Adım Adım Çözüm

1
Calculate the initial total mass of the 6 packages
Total mass=6×20=120 kg\text{Total mass} = 6 \times 20 = 120\text{ kg}
The sum of a set of numbers is equal to the product of their count and their mean.
2
Calculate the total mass after unloading the 35 kg package
New total mass=12035=85 kg\text{New total mass} = 120 - 35 = 85\text{ kg}
Removing a package decreases the total combined mass by its individual weight.
3
Calculate the new mean mass for the remaining 5 packages
New mean=855=17 kg\text{New mean} = \frac{85}{5} = 17\text{ kg}
The mean of ungrouped data is found by dividing the new sum by the updated count of items (61=56 - 1 = 5).

Anahtar Kavram

Calculation of the mean of ungrouped data after removing an item
Soru 7842Soru

A gas cylinder fitted with a pressure relief valve contains a fixed mass of an ideal gas at an initial absolute pressure of 2.50×105 Pa2.50 \times 10^5\text{ Pa} and a temperature of 27C27^\circ\text{C}. The valve is designed to open when the internal gauge pressure exceeds 3.20×105 Pa3.20 \times 10^5\text{ Pa}. Assuming the atmospheric pressure is 1.00×105 Pa1.00 \times 10^5\text{ Pa}, at what temperature will the relief valve open?

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Cevap: 231C231^\circ\text{C}

Cevap

231C231^\circ\text{C}
The correct answer is 231C231^\circ\text{C}. To solve this, first convert the initial temperature to Kelvin: T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K}. Next, calculate the final absolute pressure by adding the atmospheric pressure to the given gauge pressure: P2=3.20×105 Pa+1.00×105 Pa=4.20×105 PaP_2 = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}. Applying Gay-Lussac's Law (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}) gives T2=300×4.20×1052.50×105=504 KT_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}. Converting back to Celsius gives 504273=231C504 - 273 = 231^\circ\text{C}.

Adım Adım Çözüm

1
Convert the initial temperature from Celsius to Kelvin and determine the initial absolute pressure.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K}, P1=2.50×105 PaP_1 = 2.50 \times 10^5\text{ Pa}.
Gas laws require thermodynamic (absolute) temperature in Kelvin.
2
Calculate the final total absolute pressure (P2P_2) at which the valve opens.
P2=Pgauge+Patm=3.20×105 Pa+1.00×105 Pa=4.20×105 PaP_2 = P_{\text{gauge}} + P_{\text{atm}} = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}.
Gauge pressure measures the pressure difference relative to atmospheric pressure; gas laws use absolute pressure.
3
Apply Gay-Lussac's (Pressure) Law at constant volume to find the final Kelvin temperature (T2T_2).
P1T1=P2T2    T2=300×4.20×1052.50×105=504 K\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}.
For a fixed volume of ideal gas, absolute pressure is directly proportional to absolute temperature.
4
Convert the final temperature back to degrees Celsius.
t2=504273=231Ct_2 = 504 - 273 = 231^\circ\text{C}.
The question asks for the temperature in degrees Celsius.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law) and Absolute vs Gauge Pressure
Tahmini Süre:2m 0s
Soru 7843Soru

An autonomous drone navigating an obstacle course experiences three mutually perpendicular velocity vectors simultaneously: a horizontal forward velocity of 12 m s112\text{ m s}^{-1} due East, a horizontal crosswind drift velocity of 9 m s19\text{ m s}^{-1} due North, and a vertical downdraft velocity of 8 m s18\text{ m s}^{-1} directed straight downward. What is the magnitude of the resultant velocity vector of the drone in m s1\text{m s}^{-1}?

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Cevap: 17

Cevap

17 m s⁻¹
Because the three velocity components are mutually perpendicular, the magnitude of the overall resultant velocity is calculated using the 3D Pythagorean theorem: 122+92+82=144+81+64=289=17 m s1\sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}.

Adım Adım Çözüm

1
Identify the perpendicular vector components
vx=12 m s1v_x = 12\text{ m s}^{-1}, vy=9 m s1v_y = 9\text{ m s}^{-1}, vz=8 m s1v_z = 8\text{ m s}^{-1}
The three given velocity vectors act along mutually orthogonal spatial axes (East, North, and Downward).
2
Apply the 3D vector resultant magnitude formula
vr=vx2+vy2+vz2v_r = \sqrt{v_x^2 + v_y^2 + v_z^2}
Since the vector components are perpendicular to one another, the magnitude of their resultant is given by the extension of the Pythagorean theorem to three dimensions.
3
Substitute values and evaluate
vr=144+81+64=289=17 m s1v_r = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}
Squaring each component, adding them, and taking the principal square root yields the total magnitude of the velocity vector.

Anahtar Kavram

Magnitude of three mutually perpendicular vector components using the 3D Pythagorean theorem
Soru 7844Soru

In discharge tube experiments, electrical conduction in gases transitions through distinct physical regimes as the internal gas pressure is progressively reduced. Match each discharge phenomenon with its corresponding physical cause or operational pressure condition.

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Öğeler

Formation of the luminous positive column
Appearance and expansion of the Crookes dark space
Emission of high-velocity cathode rays
Complete cessation of electric current flow

Eşleşmeler

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Cevap

The correct pairings are: (1) Formation of the luminous positive column matches continuous de-excitation and radiative recombination of gas atoms at pressures around 1.0 mmHg1.0\text{ mmHg}; (2) Appearance and expansion of the Crookes dark space matches increase in the electron mean free path at pressures around 0.01 mmHg0.01\text{ mmHg}; (3) Emission of high-velocity cathode rays matches bombardment of the cathode by energetic positive ions releasing secondary electrons below 0.01 mmHg0.01\text{ mmHg}; and (4) Complete cessation of electric current flow matches extensive evacuation below 104 mmHg10^{-4}\text{ mmHg} leaving insufficient gas molecules for ionization.
Conduction through gases relies heavily on pressure. At moderate low pressure (1.0 mmHg1.0\text{ mmHg}), excited gas atoms emit light forming the positive column. Decreasing pressure to 0.01 mmHg0.01\text{ mmHg} increases the electron mean free path to produce the Crookes dark space and generates energetic cathode rays through ion bombardment. Extreme evacuation (<104 mmHg< 10^{-4}\text{ mmHg}) removes all gas charge carriers, halting electric conduction.

Adım Adım Çözüm

1
Analyze the pressure regime of 1.0 mmHg1.0\text{ mmHg} in a discharge tube.
Identified the positive column as the main luminous region filling most of the tube due to atom excitation and light emission.
At this pressure, gas density is sufficient to undergo repeated inelastic collisions that excite atoms and produce visible glow.
2
Examine the physical origin of the Crookes dark space at 0.01 mmHg0.01\text{ mmHg}.
Understood that lower gas density increases electron mean free path.
Electrons near the cathode travel a longer distance before hitting gas particles, creating a dark gap where collisions do not occur.
3
Determine how cathode rays are emitted at very low pressures.
Linked cathode ray emission to positive ion impact on the cathode surface.
High electric fields accelerate residual positive ions to strike the cathode, causing secondary electron emission.
4
Evaluate the extreme vacuum limit below 104 mmHg10^{-4}\text{ mmHg}.
Concluded that current stops when gas particles are virtually absent.
Gases conduct electricity via ion and electron production from collision ionization; eliminating gas molecules prevents charge transport.

Anahtar Kavram

Pressure-dependent regimes of gas conduction and cathode ray generation
Tahmini Süre:2m 0s
Soru 7845Soru

What is the simplified form of the expression 32+2332\frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqrt{2}}?

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Cevap: 12+5612 + 5\sqrt{6}

Cevap

The simplified form is 12+5612 + 5\sqrt{6}.
Multiplying both numerator and denominator by the conjugate of the denominator, (3+2)(\sqrt{3} + \sqrt{2}), clears the square roots in the denominator (resulting in 32=13 - 2 = 1). Expanding the numerator gives 36+6+6+263\sqrt{6} + 6 + 6 + 2\sqrt{6}, which simplifies cleanly to 12+5612 + 5\sqrt{6}.

Adım Adım Çözüm

1
Identify the conjugate of the denominator
The conjugate of (32)(\sqrt{3} - \sqrt{2}) is (3+2)(\sqrt{3} + \sqrt{2}).
To rationalize a binomial denominator of the form (ab)(\sqrt{a} - \sqrt{b}), multiply numerator and denominator by (a+b)(\sqrt{a} + \sqrt{b}).
2
Multiply the numerator and denominator by the conjugate
(32+23)(3+2)(32)(3+2)\frac{(3\sqrt{2} + 2\sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}
This removes the radical terms from the denominator using the difference of squares identity.
3
Expand the numerator and simplify the denominator
Denominator: (3)2(2)2=32=1(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1.
Numerator: 323+322+233+232=36+6+6+263\sqrt{2}\cdot\sqrt{3} + 3\sqrt{2}\cdot\sqrt{2} + 2\sqrt{3}\cdot\sqrt{3} + 2\sqrt{3}\cdot\sqrt{2} = 3\sqrt{6} + 6 + 6 + 2\sqrt{6}.
Apply the distributive property and basic radical simplification rules.
4
Combine like terms in the numerator
(6+6)+(36+26)=12+56(6 + 6) + (3\sqrt{6} + 2\sqrt{6}) = 12 + 5\sqrt{6}.
Collect rational numbers together and like surd terms together.

Anahtar Kavram

Rationalization of Denominators with Binomial Surds
Soru 7846Soru

If xx and yy are real numbers satisfying the simultaneous equations xy=4x - y = 4 and x2+y2=26x^2 + y^2 = 26, what is the value of the product xyxy?

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Cevap: 5

Cevap

The value of the product xyxy is 5.
Expressing xx from the linear equation yields x=y+4x = y + 4. Substituting this into x2+y2=26x^2 + y^2 = 26 gives (y+4)2+y2=26(y + 4)^2 + y^2 = 26, which expands and simplifies to y2+4y5=0y^2 + 4y - 5 = 0. Factoring gives solutions y=1y = 1 (with x=5x = 5) and y=5y = -5 (with x=1x = -1). Both solution pairs (5,1)(5, 1) and (1,5)(-1, -5) result in the product xy=5xy = 5.

Adım Adım Çözüm

1
Express xx in terms of yy using the linear equation
x=y+4x = y + 4
Isolate xx to substitute into the second equation.
2
Substitute x=y+4x = y + 4 into the quadratic equation x2+y2=26x^2 + y^2 = 26
(y+4)2+y2=26(y + 4)^2 + y^2 = 26
Reduce the system to a single quadratic equation in yy.
3
Expand and simplify into standard quadratic form
y2+4y5=0y^2 + 4y - 5 = 0
Expanding gives 2y2+8y+16=262y^2 + 8y + 16 = 26, which simplifies by subtracting 26 and dividing by 2.
4
Solve for yy by factoring
y=1y = 1 or y=5y = -5
The factors of y2+4y5y^2 + 4y - 5 are (y+5)(y1)=0(y + 5)(y - 1) = 0.
5
Compute corresponding xx values and the product xyxy
For y=1y = 1, x=5    xy=5x = 5 \implies xy = 5; for y=5y = -5, x=1    xy=5x = -1 \implies xy = 5
Substitute each yy back into x=y+4x = y + 4 and evaluate xyxy.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
Soru 7847Soru

A cannonball is fired from level ground with an initial speed of 20 m/s20\text{ m/s} at an angle of 3030^\circ above the horizontal. Taking the acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the total time of flight of the cannonball before it returns to ground level?

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Cevap: 2.0 s2.0\text{ s}

Cevap

2.0 s2.0\text{ s}
The correct answer of 2.0 s2.0\text{ s} is determined by resolving the initial speed into its vertical component uy=20sin30=10 m/su_y = 20 \sin 30^\circ = 10\text{ m/s} and using the total time of flight formula T=2usinθg=2(10)10=2.0 sT = \frac{2 u \sin \theta}{g} = \frac{2(10)}{10} = 2.0\text{ s}.

Adım Adım Çözüm

1
Calculate the vertical component of the initial velocity (uyu_y).
uy=usinθ=20×sin30=20×0.5=10 m/su_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\text{ m/s}
Only the vertical component of velocity determines the time the projectile remains in the air.
2
Calculate the total time of flight (TT) using the kinematic formula for full trajectory.
T=2uyg=2×1010=2.0 sT = \frac{2 u_y}{g} = \frac{2 \times 10}{10} = 2.0\text{ s}
The total time of flight includes both the time to ascend to peak height and descend back to the launch height under gravity gg.

Anahtar Kavram

Total Time of Flight in Projectile Motion
Soru 7848Soru

The table below shows the distribution of examination scores of 5050 candidates in a selection test:

Score ClassFrequency (ff)
101910 - 1955
202920 - 2999
303930 - 391616
404940 - 491414
505950 - 5966

Using the cumulative frequency distribution (ogive), what is the estimated 70th70^{\text{th}} percentile score?

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Cevap: 43.143.1

Cevap

The estimated 70th70^{\text{th}} percentile score is 43.143.1.
The 70th70^{\text{th}} percentile corresponds to the score at the 35th35^{\text{th}} candidate (70%70\% of 5050). This falls within the 404940 - 49 score class, which has a lower boundary of 39.539.5, a frequency of 1414, and a class width of 1010. Interpolating linearly gives 39.5+(353014)×10=43.0743.139.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 43.07 \approx 43.1.

Adım Adım Çözüm

1
Construct the cumulative frequency table to find class boundaries and cumulative frequencies.
Cumulative frequencies: 101910-19 (CF=5CF = 5), 202920-29 (CF=14CF = 14), 303930-39 (CF=30CF = 30), 404940-49 (CF=44CF = 44), 505950-59 (CF=50CF = 50).
Cumulative frequencies are required to locate the position of the desired percentile.
2
Determine the rank position of the 70th70^{\text{th}} percentile (P70P_{70}).
Rank position = 70100×50=35th\frac{70}{100} \times 50 = 35^{\text{th}} position.
The 70th70^{\text{th}} percentile corresponds to the score below which 70%70\% of the total candidates fall.
3
Identify the class interval containing the 35th35^{\text{th}} cumulative frequency and state its parameters.
The target class is 404940 - 49. Parameters: Lower class boundary L=39.5L = 39.5, preceding cumulative frequency CFprev=30CF_{\text{prev}} = 30, class frequency f=14f = 14, class width c=10c = 10.
Since 30<354430 < 35 \le 44, the 35th35^{\text{th}} entry falls within the 404940 - 49 class.
4
Apply the linear interpolation formula for percentiles from an ogive.
P70=L+(70N100CFprevf)×c=39.5+(353014)×10=39.5+501443.0743.1P_{70} = L + \left(\frac{\frac{70N}{100} - CF_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 39.5 + \frac{50}{14} \approx 43.07 \approx 43.1.
This calculates the exact score estimate on the cumulative frequency curve.

Anahtar Kavram

Estimating Percentiles from Grouped Data and Ogives
Soru 7849Soru

An electron inside an excited atom drops from an energy state of 2.10 eV-2.10\text{ eV} to a lower energy state of 4.575 eV-4.575\text{ eV}. What is the wavelength of the emitted photon in nanometers (nm\text{nm})? (Take Planck's constant h=6.6×1034 J sh = 6.6 \times 10^{-34}\text{ J s}, speed of light c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, and 1 eV=1.6×1019 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}).

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Cevap: 500

Cevap

The wavelength of the emitted photon is 500 nm500\text{ nm}.
The energy of the emitted photon is calculated from the energy change of the electron, ΔE=2.10 eV(4.575 eV)=2.475 eV\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}. Converting this to Joules gives 2.475×1.6×1019 J=3.96×1019 J2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}. Substituting into the wavelength formula λ=hcΔE=6.6×1034×3.0×1083.96×1019=5.0×107 m=500 nm\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}.

Adım Adım Çözüm

1
Calculate the energy difference between the initial excited state and final state
ΔE=2.10 eV(4.575 eV)=2.475 eV\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}
The energy of the emitted photon equals the difference between the two energy levels.
2
Convert the photon energy from electron-volts (eV) to Joules (J)
ΔE=2.475×1.6×1019 J=3.96×1019 J\Delta E = 2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}
Standard SI units must be used to calculate wavelength in meters.
3
Apply the Planck-Einstein relation λ=hcΔE\lambda = \frac{hc}{\Delta E} to calculate the wavelength in meters and convert to nanometers
λ=(6.6×1034 J s)(3.0×108 m/s)3.96×1019 J=5.0×107 m=500 nm\lambda = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m/s})}{3.96 \times 10^{-19}\text{ J}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}
Converting meters to nanometers requires multiplying by 10910^9.

Anahtar Kavram

Photon emission wavelength during atomic energy level transitions
Tahmini Süre:1m 30s
Soru 7850Soru

Given the universal set U={xZ:1x20}U = \{x \in \mathbb{Z} : 1 \le x \le 20\}, with subsets P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\} and Q={xU:x is an odd integer}Q = \{x \in U : x \text{ is an odd integer}\}, what is the cardinal number of (PQ)(P \cup Q)'?

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Cevap: 99

Cevap

99
The universal set contains 2020 elements. The union PQP \cup Q consists of all odd numbers and prime numbers between 11 and 2020, giving 1111 unique elements: {1,2,3,5,7,9,11,13,15,17,19}\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}. Subtracting these 1111 elements from the total 2020 elements in UU gives n((PQ))=2011=9n((P \cup Q)') = 20 - 11 = 9.

Adım Adım Çözüm

1
Identify the elements of the universal set UU and subsets PP and QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, so n(U)=20n(U) = 20.
P={2,3,5,7,11,13,17,19}P = \{2, 3, 5, 7, 11, 13, 17, 19\}
Q={1,3,5,7,9,11,13,15,17,19}Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}
Explicitly listing the set elements helps correctly calculate the union.
2
Determine the union PQP \cup Q.
PQ={1,2,3,5,7,9,11,13,15,17,19}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, which contains 1111 elements.
The union combines all unique elements present in set PP, set QQ, or both.
3
Calculate the complement set (PQ)(P \cup Q)' and its cardinality.
(PQ)=U(PQ)={4,6,8,10,12,14,16,18,20}(P \cup Q)' = U \setminus (P \cup Q) = \{4, 6, 8, 10, 12, 14, 16, 18, 20\}, so n((PQ))=2011=9n((P \cup Q)') = 20 - 11 = 9.
The complement of a set contains all elements in the universal set UU that are not in the given set.

Anahtar Kavram

Set Operations and Complement of Sets
Tahmini Süre:1m 0s
Soru 7851Soru

A micrometer screw gauge with a pitch of 0.5 mm0.5\text{ mm} and 5050 divisions on its circular thimble scale has a negative zero error such that when the jaws are fully closed, the 46th46\text{th} division on the thimble scale aligns with the main scale datum line. When this instrument is used to measure the total thickness of a stack of 4040 identical sheets of paper, the main scale reads 10.0 mm10.0\text{ mm} and the thimble scale reads 1616 divisions. What is the actual thickness of a single sheet of paper?

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Cevap: 0.255 mm0.255\text{ mm}

Cevap

The actual thickness of a single sheet of paper is 0.255 mm0.255\text{ mm}.
The least count of the micrometer is 0.5 mm50=0.01 mm\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}. Because the 46th division coincides with the datum line upon closure, the zero error is negative: (5046)×0.01 mm=0.04 mm-(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}. The observed reading is 10.0 mm+0.16 mm=10.16 mm10.0\text{ mm} + 0.16\text{ mm} = 10.16\text{ mm}. Correcting for zero error gives a true total thickness of 10.16 mm(0.04 mm)=10.20 mm10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}. Dividing this total thickness by 40 sheets gives an individual sheet thickness of 0.255 mm0.255\text{ mm}.

Adım Adım Çözüm

1
Determine the least count (LC) of the micrometer screw gauge.
LC=PitchNumber of thimble divisions=0.5 mm50=0.01 mm\text{LC} = \frac{\text{Pitch}}{\text{Number of thimble divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}.
The least count is the minimum distance measurable by one circular scale division.
2
Calculate the zero error of the instrument.
Zero error=(5046)×0.01 mm=0.04 mm\text{Zero error} = -(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}.
Since the 46th division aligns with the datum line when jaws are closed, the zero mark lies 4 divisions below the line, indicating a negative zero error.
3
Find the observed reading for the stack of 40 sheets.
Observed reading=Main scale+(Thimble reading×LC)=10.0 mm+(16×0.01 mm)=10.16 mm\text{Observed reading} = \text{Main scale} + (\text{Thimble reading} \times \text{LC}) = 10.0\text{ mm} + (16 \times 0.01\text{ mm}) = 10.16\text{ mm}.
The total observed value combines the main scale reading and the fractional thimble scale reading.
4
Calculate the true (corrected) total thickness of the stack.
True reading=Observed readingZero error=10.16 mm(0.04 mm)=10.20 mm\text{True reading} = \text{Observed reading} - \text{Zero error} = 10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}.
Zero error correction requires subtracting the zero error (with sign) from the observed reading.
5
Calculate the thickness of a single sheet of paper.
Thickness of one sheet=10.20 mm40=0.255 mm\text{Thickness of one sheet} = \frac{10.20\text{ mm}}{40} = 0.255\text{ mm}.
Dividing the total corrected thickness by the total number of sheets gives the thickness per sheet.

Anahtar Kavram

Measurement of length using micrometer screw gauge with negative zero error correction
Tahmini Süre:2m 0s
Soru 7852Soru

A periodic water wave of frequency 5.0 Hz5.0\text{ Hz} has a wavelength of 1.2 m1.2\text{ m} in deep water. When the wave enters a shallow section of the ripple tank, its speed reduces to 4.0 m s14.0\text{ m s}^{-1}. What is the wavelength of the wave in the shallow section?

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Cevap: 0.80 m0.80\text{ m}

Cevap

0.80 m0.80\text{ m}
When a wave travels from one medium to another (e.g., deep to shallow water), its frequency ff remains constant because frequency is dependent only on the wave source. Using the wave equation v=fλv = f\lambda, the wavelength in the shallow water is calculated as λ=vf=4.0 m s15.0 Hz=0.80 m\lambda = \frac{v}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}.

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1
Determine the invariant property across media boundaries
The frequency ff remains constant at 5.0 Hz5.0\text{ Hz} when a wave passes from deep water to shallow water.
Frequency is determined solely by the source of the wave vibration, not the medium of propagation.
2
Calculate the wavelength in the shallow section using the wave equation
λ2=v2f=4.0 m s15.0 Hz=0.80 m\lambda_2 = \frac{v_2}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}
Applying the relationship v=fλv = f\lambda for the second medium.

Anahtar Kavram

Frequency invariance of waves across boundaries and application of the wave equation v=fλv = f\lambda
Soru 7853Soru

X-rays are high-frequency electromagnetic waves that undergo deflection when passing through strong electric or magnetic fields.

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Cevap: False

Cevap

The statement is False because X-rays carry no electrical charge and therefore pass through electric and magnetic fields without deflection.
X-rays are electromagnetic radiation made of photons with zero electric charge. Because they carry no charge, they travel in straight lines and remain entirely unaffected by surrounding electric or magnetic fields.

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1
Identify the physical nature of X-rays.
X-rays are electromagnetic radiation (photons) of high frequency and short wavelength.
Electromagnetic waves travel at the speed of light and consist of transverse oscillating electric and magnetic fields, carrying energy but no electric charge.
2
Analyze the effect of electric and magnetic fields on uncharged radiation.
The electric force FE=qEF_E = qE and magnetic force FB=qvBsinθF_B = qvB sin\theta are both zero when q=0q = 0.
Since photons have zero net electrical charge (q=0q = 0), they experience no net deflecting force in external electric or magnetic fields.

Anahtar Kavram

Neutrality of X-rays in electromagnetic fields
Soru 7854Soru

If θ\theta is an acute angle such that sinθ=513\sin \theta = \frac{5}{13}, what is the exact numerical value of 169(sin2θcos2θ)169(\sin^2 \theta - \cos^2 \theta)?

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Cevap: -119

Cevap

The exact numerical value of the expression is 119-119.
Using the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, we find cos2θ=125169=144169\cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}. Then sin2θcos2θ=25144169=119169\sin^2 \theta - \cos^2 \theta = \frac{25 - 144}{169} = -\frac{119}{169}. Multiplying by 169169 gives 119-119.

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1
Calculate cos2θ\cos^2 \theta using the Pythagorean trigonometric identity.
cos2θ=144169\cos^2 \theta = \frac{144}{169}
Since sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, subtracting sin2θ=25169\sin^2 \theta = \frac{25}{169} from 11 yields 144169\frac{144}{169}.
2
Compute the difference sin2θcos2θ\sin^2 \theta - \cos^2 \theta.
sin2θcos2θ=119169\sin^2 \theta - \cos^2 \theta = -\frac{119}{169}
Subtracting 144169\frac{144}{169} from 25169\frac{25}{169} gives 119169-\frac{119}{169}.
3
Scale the difference by 169169.
169×(119169)=119169 \times \left(-\frac{119}{169}\right) = -119
Multiplying 119169-\frac{119}{169} by 169169 cancels the denominator, leaving 119-119.

Anahtar Kavram

Pythagorean Trigonometric Identity
Soru 7855Soru

A girl stands at a specific distance from a flat vertical wall and claps her hands once. If she hears the echo 0.4 s0.4\text{ s} later, what is her distance from the wall in meters? (Take the speed of sound in air as 340 m/s340\text{ m/s})

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Cevap: 68

Cevap

The distance of the girl from the wall is 68 m68\text{ m}.
An echo involves the sound traveling to the reflecting surface and returning to the source, covering a total distance of 2d2d. Using 2d=v×t2d = v \times t, we obtain d=340×0.42=68 md = \frac{340 \times 0.4}{2} = 68\text{ m}.

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1
Identify the given values from the problem statement.
Time for echo t=0.4 st = 0.4\text{ s}, speed of sound v=340 m/sv = 340\text{ m/s}.
An echo is a reflected sound wave that travels to the wall and back, completing a round trip.
2
Apply the echo calculation formula.
d=v×t2d = \frac{v \times t}{2}
The total distance traveled by the sound is 2d2d. Thus, the one-way distance dd to the reflecting surface is half of the total distance.
3
Substitute the values and calculate the distance.
d=340×0.42=68 md = \frac{340 \times 0.4}{2} = 68\text{ m}
Multiplying the speed by half the elapsed time gives the distance to the wall.

Anahtar Kavram

Echo distance calculation
Soru 7856Soru

Given the simultaneous equations xy=1x - y = 1 and x2+y2=25x^2 + y^2 = 25, where xx and yy are both positive real numbers, what is the value of x+yx + y?

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Cevap: 77

Cevap

The value of x+yx + y is 77.
Expressing xx as y+1y + 1 and substituting into x2+y2=25x^2 + y^2 = 25 yields 2y2+2y24=02y^2 + 2y - 24 = 0. Factoring gives y=3y = 3 (rejecting y=4y = -4 as y>0y > 0). Substituting y=3y = 3 back gives x=4x = 4. Adding these values together yields x+y=7x + y = 7.

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1
Express xx in terms of yy using the linear equation.
x=y+1x = y + 1
Rearranging xy=1x - y = 1 allows substitution into the quadratic equation.
2
Substitute x=y+1x = y + 1 into the quadratic equation x2+y2=25x^2 + y^2 = 25.
(y+1)2+y2=25    y2+2y+1+y2=25    2y2+2y24=0(y + 1)^2 + y^2 = 25 \implies y^2 + 2y + 1 + y^2 = 25 \implies 2y^2 + 2y - 24 = 0
This reduces the system to a single quadratic equation in terms of yy.
3
Solve the quadratic equation for yy.
y2+y12=0    (y+4)(y3)=0    y=3y^2 + y - 12 = 0 \implies (y + 4)(y - 3) = 0 \implies y = 3 or y=4y = -4
Dividing by 22 simplifies the equation, and factoring gives the potential roots for yy.
4
Select the positive value of yy and calculate xx and x+yx + y.
Since y>0y > 0, y=3y = 3. Then x=3+1=4x = 3 + 1 = 4, so x+y=4+3=7x + y = 4 + 3 = 7.
The question specifies that both xx and yy are positive real numbers.

Anahtar Kavram

Solving Simultaneous Linear and Quadratic Equations by Substitution
Soru 7857Soru

An electron beam traveling horizontally enters a region containing mutually perpendicular uniform electric and magnetic fields. The electric field intensity is 4.4×103 V m14.4 \times 10^3\text{ V m}^{-1} and the magnetic flux density is 5.0×104 T5.0 \times 10^{-4}\text{ T}. If the beam passes through undeflected and subsequently enters a region containing only the magnetic field BB, what is the radius of the circular path executed by the electrons? (Take the specific charge of an electron em=1.76×1011 C kg1\frac{e}{m} = 1.76 \times 10^{11}\text{ C kg}^{-1})

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Cevap: 0.10 m0.10\text{ m}

Cevap

The radius of the circular path executed by the electrons in the magnetic field is 0.10 m0.10\text{ m}.
Under velocity selection conditions, equal electric and magnetic forces (eE=evBeE = evB) establish electron speed v=EB=8.8×106 m s1v = \frac{E}{B} = 8.8 \times 10^6\text{ m s}^{-1}. When moving solely through magnetic field BB, magnetic force provides centripetal force (evB=mv2revB = \frac{mv^2}{r}), yielding orbital radius r=v(e/m)B=8.8×1061.76×1011×5.0×104=0.10 mr = \frac{v}{(e/m)B} = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = 0.10\text{ m}.

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1
Calculate the electron velocity using the undeflected crossed fields condition (velocity selector)
v=EB=4.4×103 V m15.0×104 T=8.8×106 m s1v = \frac{E}{B} = \frac{4.4 \times 10^3\text{ V m}^{-1}}{5.0 \times 10^{-4}\text{ T}} = 8.8 \times 10^6\text{ m s}^{-1}
When electric and magnetic forces are equal and opposite (eE=evBeE = evB), the electrons travel in a straight line undeflected.
2
Equate magnetic force to centripetal force in the region with magnetic field only
evB=mv2r    r=mveB=v(em)BevB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} = \frac{v}{\left(\frac{e}{m}\right)B}
The magnetic Lorentz force provides the necessary centripetal force for circular motion.
3
Substitute numerical values to find the radius rr
r=8.8×1061.76×1011×5.0×104=8.8×1068.8×107=0.10 mr = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = \frac{8.8 \times 10^6}{8.8 \times 10^7} = 0.10\text{ m}
Direct algebraic simplification yields the orbital radius in meters.

Anahtar Kavram

Deflection of cathode rays (electrons) in crossed electric and magnetic fields (velocity selector) and circular orbital dynamics in uniform magnetic fields.
Tahmini Süre:2m 0s
Soru 7858Soru

The table below shows the distribution of weekly overtime hours worked by 1010 technicians in a manufacturing plant:

Overtime Hours (xx)2468
Number of Technicians (ff)4321

What is the variance of the overtime hours?

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Cevap: 4

Cevap

The variance of the overtime hours is 44.
The correct answer is 44. First, compute the mean xˉ=4010=4\bar{x} = \frac{40}{10} = 4. Then compute the sum of weighted squared deviations f(xxˉ)2=4(24)2+3(44)2+2(64)2+1(84)2=16+0+8+16=40\sum f(x - \bar{x})^2 = 4(2-4)^2 + 3(4-4)^2 + 2(6-4)^2 + 1(8-4)^2 = 16 + 0 + 8 + 16 = 40. Dividing this by total frequency f=10\sum f = 10 yields σ2=4010=4\sigma^2 = \frac{40}{10} = 4.

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1
Calculate the total frequency (f\sum f) and the sum of the products of values and frequencies (fx\sum fx).
f=4+3+2+1=10\sum f = 4 + 3 + 2 + 1 = 10 and fx=(2×4)+(4×3)+(6×2)+(8×1)=8+12+12+8=40\sum fx = (2 \times 4) + (4 \times 3) + (6 \times 2) + (8 \times 1) = 8 + 12 + 12 + 8 = 40.
These sums are required to compute the mean of the distribution.
2
Determine the mean (xˉ\bar{x}) of the distribution.
xˉ=fxf=4010=4\bar{x} = \frac{\sum fx}{\sum f} = \frac{40}{10} = 4.
The mean is used as the central reference point to find deviations.
3
Find the squared deviation (xxˉ)2(x - \bar{x})^2 for each score and multiply by its respective frequency ff.
For x=2x = 2: 4(24)2=164(2 - 4)^2 = 16.
For x=4x = 4: 3(44)2=03(4 - 4)^2 = 0.
For x=6x = 6: 2(64)2=82(6 - 4)^2 = 8.
For x=8x = 8: 1(84)2=161(8 - 4)^2 = 16.
Total sum f(xxˉ)2=16+0+8+16=40\sum f(x - \bar{x})^2 = 16 + 0 + 8 + 16 = 40.
Variance evaluates the average of these weighted squared deviations.
4
Compute the variance (σ2\sigma^2) by dividing the sum of weighted squared deviations by the total frequency.
σ2=f(xxˉ)2f=4010=4\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{40}{10} = 4.
This gives the population variance for the given frequency distribution.

Anahtar Kavram

Variance for Frequency Distributions
Tahmini Süre:1m 30s
Soru 7859Soru

A trapezium has an area of 180 cm2180\text{ cm}^2 and a perpendicular height of 12 cm12\text{ cm}. If the lengths of its two parallel sides are in the ratio 2:32:3, calculate the length, in cm\text{cm}, of the longer parallel side.

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Cevap: 18

Cevap

The length of the longer parallel side is 18 cm18\text{ cm}.
By representing the parallel sides as 2x2x and 3x3x, the trapezium area formula A=12(a+b)hA = \frac{1}{2}(a + b)h gives 180=12(2x+3x)(12)=30x180 = \frac{1}{2}(2x + 3x)(12) = 30x. Solving for xx yields x=6x = 6. Therefore, the longer parallel side is 3(6)=18 cm3(6) = 18\text{ cm}.

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1
Express the parallel sides algebraically from the ratio 2:32:3.
Let the shorter parallel side a=2xa = 2x and the longer parallel side b=3xb = 3x.
Using a common multiplier xx preserves the given side ratio.
2
Substitute the expressions and known values into the trapezium area formula.
180=12(2x+3x)×12180 = \frac{1}{2}(2x + 3x) \times 12
The area AA of a trapezium is given by A=12(a+b)hA = \frac{1}{2}(a + b)h.
3
Solve for the variable xx.
180=6×5x    30x=180    x=6180 = 6 \times 5x \implies 30x = 180 \implies x = 6
Simplifying 12×12=6\frac{1}{2} \times 12 = 6 and multiplying by (2x+3x)=5x(2x + 3x) = 5x.
4
Calculate the length of the longer parallel side.
Longer side =3x=3×6=18 cm= 3x = 3 \times 6 = 18\text{ cm}
The longer side corresponds to the 3x3x term in the ratio.

Anahtar Kavram

Area of a trapezium involving algebraic ratio problem solving
Soru 7860Soru

A Vernier caliper with a least count of 0.01 cm0.01\text{ cm} is used to determine the outer diameter of a cylindrical tube. When the measuring jaws are brought together without any object between them, the zero mark of the vernier scale lies to the right of the main scale zero mark, with the 3rd3\text{rd} vernier division coinciding with a main scale line. When measuring the tube, the main scale reads 4.20 cm4.20\text{ cm} and the 6th6\text{th} vernier division coincides with a main scale line. What is the actual outer diameter of the tube?

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Cevap: 4.23 cm4.23\text{ cm}

Cevap

The actual outer diameter of the tube is 4.23 cm4.23\text{ cm}.
The observed reading of the Vernier caliper is calculated as Main Scale Reading + (Coinciding Division × Least Count) = 4.20 cm+(6×0.01 cm)=4.26 cm4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}. Because the vernier zero lies to the right when closed, it has a positive zero error of +0.03 cm+0.03\text{ cm}. Subtracting this zero error from the observed reading gives the true measurement of 4.23 cm4.23\text{ cm}.

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1
Determine the zero error of the Vernier caliper
Zero Error =+(3×0.01 cm)=+0.03 cm= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}
Since the zero of the vernier scale lies to the right of the main scale zero, the instrument has a positive zero error.
2
Calculate the observed reading
Observed Reading =4.20 cm+(6×0.01 cm)=4.26 cm= 4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}
The observed measurement is the main scale reading plus the product of the coinciding vernier division and the least count.
3
Calculate the actual (corrected) reading
Actual Reading =Observed ReadingZero Error=4.26 cm(+0.03 cm)=4.23 cm= \text{Observed Reading} - \text{Zero Error} = 4.26\text{ cm} - (+0.03\text{ cm}) = 4.23\text{ cm}
Zero error must always be subtracted algebraically from the observed reading to obtain the accurate measurement.

Anahtar Kavram

Zero Error Correction in Vernier Calipers
Tahmini Süre:1m 30s
ÖncekiSayfa 393 / 697Sonraki
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