Tüm alıştırma soruları

13931 soru

Soru 4401Soru

A cooperative society invested 80,000\text{₦}80,000 in a fixed deposit fund that pays compound interest at a rate of 10%10\% per annum compounded annually. What is the total compound interest earned by the cooperative society at the end of 22 years?

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Cevap: 16,800\text{₦}16,800

Cevap

16,800\text{₦}16,800
Using the compound interest formula for total amount, A=P(1+r100)nA = P\left(1 + \frac{r}{100}\right)^n, with principal P=80,000P = \text{₦}80,000, rate r=10%r = 10\%, and time n=2n = 2 years: A=80000×(1.1)2=96,800A = 80000 \times (1.1)^2 = \text{₦}96,800. The interest earned is the difference between the total amount and the principal: I=9680080000=16,800I = 96800 - 80000 = \text{₦}16,800.

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1
Calculate the total accumulated amount AA using the compound interest formula A=P(1+r100)nA = P\left(1 + \frac{r}{100}\right)^n
A=80000×(1+10100)2=80000×(1.1)2=80000×1.21=96,800A = 80000 \times \left(1 + \frac{10}{100}\right)^2 = 80000 \times (1.1)^2 = 80000 \times 1.21 = \text{₦}96,800
The formula yields the full final balance after compound growth over 2 years.
2
Subtract the initial principal PP from the total amount AA to find the compound interest earned II
I=AP=9680080000=16,800I = A - P = 96800 - 80000 = \text{₦}16,800
Interest earned represents only the financial growth beyond the starting capital.

Anahtar Kavram

Compound Interest and Total Accumulated Amount
Soru 4402Soru

Complete the sentence below by filling in the blank with the appropriate specialized medical term.

Aşağıdaki boşlukları doldurun

During a routine cardiovascular examination, the physician performed to listen to the internal sounds of the patient's heart and lungs using a stethoscope.
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Cevap

auscultation
The word 'auscultation' is the precise medical register term for the act of listening to internal body sounds, particularly using a stethoscope to examine cardiac and respiratory functions.

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1
Analyze the context of the sentence to determine the relevant professional domain and action.
The setting is medical clinical examination, specifically the act of listening to internal bodily organs like the heart and lungs using a stethoscope.
Specialized registers require choosing the precise technical word assigned to a specific professional procedure.
2
Identify the precise medical register term for listening to organ sounds.
The correct technical term is 'auscultation'.
'Auscultation' is derived from Latin meaning 'to listen', and specifically denotes diagnostic listening to body sounds.

Anahtar Kavram

Occupational and Specialized Registers: Medical Terminology
Soru 4403Soru

A container holds a large number of red, blue, and yellow counters. The theoretical probability of selecting a red counter at random is 25\frac{2}{5}, and the theoretical probability of selecting a blue counter is 13\frac{1}{3}. In an experiment where a counter is drawn and replaced 300300 times, a yellow counter is selected 7272 times. What is the absolute difference between the theoretical expected number of yellow counters and the experimental frequency of yellow counters obtained?

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Cevap: 88

Cevap

The absolute difference between the theoretical expected number of yellow counters and the experimental frequency is 88.
The theoretical probability of drawing a yellow counter is 1(25+13)=4151 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15}. For 300300 trials, the theoretical expected count of yellow counters is 415×300=80\frac{4}{15} \times 300 = 80. Subtracting the experimental observed count of 7272 from 8080 gives an absolute difference of 88.

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1
Calculate the theoretical probability of drawing a yellow counter.
P(Yellow)=1(P(Red)+P(Blue))=1(25+13)=11115=415P(\text{Yellow}) = 1 - \left(P(\text{Red}) + P(\text{Blue})\right) = 1 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}
The sum of probabilities of all mutually exclusive outcomes in the sample space must equal 11.
2
Calculate the theoretical expected frequency of yellow counters in 300300 trials.
Expected frequency=P(Yellow)×Total trials=415×300=80\text{Expected frequency} = P(\text{Yellow}) \times \text{Total trials} = \frac{4}{15} \times 300 = 80
The expected number of occurrences of an outcome is the product of its theoretical probability and the number of trials.
3
Find the absolute difference between expected frequency and experimental frequency.
|80 - 72| = 8
The question asks for the difference between the theoretical expected value (8080) and the actual experimental result (7272).

Anahtar Kavram

Experimental and Theoretical Probability
Tahmini Süre:1m 30s
Soru 4404Soru

A rectangular wire loop is pulled horizontally to the right out of a region containing a uniform magnetic field directed perpendicularly into the page. According to Lenz's law, what is the direction of the induced current in the loop and the direction of the resulting magnetic force acting on the loop?

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Cevap: Clockwise induced current; magnetic force directed to the left

Cevap

Clockwise induced current; magnetic force directed to the left
According to Lenz's law, an induced electric current flows in a direction such that its magnetic field opposes the change in magnetic flux that produced it. As the loop is pulled to the right, the magnetic flux pointing into the page decreases. The loop responds by inducing a current that creates additional magnetic field into the page to resist this decrease. By the right-hand rule, a current circulating clockwise produces a magnetic field into the page. Additionally, the magnetic force created on the loop must oppose the external motion, acting to the left.

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1
Determine the change in magnetic flux passing through the loop
As the rectangular loop is pulled to the right out of the magnetic field region, the magnetic flux directed into the page through the loop is decreasing.
Electromagnetic induction depends on the rate of change of magnetic flux.
2
Apply Lenz's law to determine the direction of the induced magnetic field and current
To oppose the decrease in inward flux, the induced current must create its own magnetic field directed into the page. By the right-hand grip rule, an inward induced field corresponds to a clockwise current flow.
Lenz's law states that the direction of an induced current always opposes the change in magnetic flux causing it.
3
Determine the direction of the net magnetic force on the loop
The induced magnetic force must oppose the mechanical motion pulling the loop to the right, so the net magnetic force acts to the left.
Lenz's law is a consequence of the conservation of energy, ensuring mechanical work must be done against electromagnetic forces.

Anahtar Kavram

Lenz's Law and Direction of Induced Current
Soru 4405Soru

A rigid vessel of fixed volume contains an ideal gas at an initial pressure of 1.50×105 Pa1.50 \times 10^5\text{ Pa} and a temperature of 27C27^\circ\text{C}. Additional gas is pumped into the vessel until the total number of moles of gas is doubled. If the temperature of the gas increases to 87C87^\circ\text{C}, what is the final pressure of the gas in units of 105 Pa10^5\text{ Pa}?

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Cevap: 3.6

Cevap

The final pressure of the gas is 3.6×105 Pa3.6 \times 10^5\text{ Pa} (which is 3.63.6 in units of 105 Pa10^5\text{ Pa}).
Using the ideal gas equation PV=nRTPV = nRT at fixed volume, the ratio of final to initial pressure is given by P2/P1=(n2/n1)×(T2/T1)P_2/P_1 = (n_2/n_1) \times (T_2/T_1). Converting temperatures to Kelvin yields T1=300 KT_1 = 300\text{ K} and T2=360 KT_2 = 360\text{ K}. Given that the number of moles doubles (n2/n1=2n_2/n_1 = 2), substituting the values gives P2=1.50×105×2×(360/300)=3.60×105 PaP_2 = 1.50 \times 10^5 \times 2 \times (360/300) = 3.60 \times 10^5\text{ Pa}.

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1
Convert initial and final temperatures from Celsius to absolute temperature in Kelvin.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=87+273=360 KT_2 = 87 + 273 = 360\text{ K}.
Gas laws require absolute temperatures in Kelvin for thermodynamic calculations.
2
Formulate the pressure relation from the ideal gas equation PV=nRTPV = nRT.
Since volume VV and the universal gas constant RR are constant, P2=P1×n2n1×T2T1P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1}.
Pressure is directly proportional to both the number of moles and the absolute temperature when volume is fixed.
3
Substitute the mole ratio n2/n1=2n_2/n_1 = 2, initial pressure, and Kelvin temperatures to compute P2P_2.
P2=1.50×105×2×360300=3.60×105 PaP_2 = 1.50 \times 10^5 \times 2 \times \frac{360}{300} = 3.60 \times 10^5\text{ Pa}.
Evaluates the final gas pressure in the requested numerical units.

Anahtar Kavram

Ideal Gas Equation and Variable Moles under Constant Volume
Soru 4406Soru

An electric heater rated at 690 W690\text{ W} is immersed in a thermally insulated container holding a mixture of 0.50 kg0.50\text{ kg} of ice and 0.50 kg0.50\text{ kg} of liquid water in equilibrium at 0C0^\circ\text{C}. The heater is operated for 5.0 minutes5.0\text{ minutes}. Assuming negligible heat capacity for the container and no heat loss to the surroundings, what is the final equilibrium temperature of the mixture?
(Specific latent heat of fusion of ice Lf=3.3×105 J kg1L_f = 3.3 \times 10^5\text{ J kg}^{-1}, specific heat capacity of water cw=4200 J kg1 K1c_w = 4200\text{ J kg}^{-1}\text{ K}^{-1})

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Cevap: 10C10^\circ\text{C}

Cevap

The final equilibrium temperature of the mixture is 10C10^\circ\text{C}.
The total energy supplied by the 690 W690\text{ W} heater in 300 s300\text{ s} is 207,000 J207,000\text{ J}. Melting all 0.50 kg0.50\text{ kg} of ice at 0C0^\circ\text{C} requires 165,000 J165,000\text{ J}. The remaining 42,000 J42,000\text{ J} heats the combined 1.00 kg1.00\text{ kg} of water (initial water + melted ice) through ΔT=42,0001.00×4200=10C\Delta T = \frac{42,000}{1.00 \times 4200} = 10^\circ\text{C}, reaching a final equilibrium temperature of 10C10^\circ\text{C}.

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1
Calculate the total thermal energy provided by the electric heater.
Qtotal=P×t=690 W×(5.0×60 s)=207,000 JQ_{\text{total}} = P \times t = 690\text{ W} \times (5.0 \times 60\text{ s}) = 207,000\text{ J}.
Power multiplied by time yields total energy supplied.
2
Determine the energy required to completely melt the 0.50 kg0.50\text{ kg} of ice at 0C0^\circ\text{C}.
Qmelt=mice×Lf=0.50 kg×330,000 J kg1=165,000 JQ_{\text{melt}} = m_{\text{ice}} \times L_f = 0.50\text{ kg} \times 330,000\text{ J kg}^{-1} = 165,000\text{ J}.
Latent heat of fusion changes state from solid ice to liquid water at constant temperature 0C0^\circ\text{C}.
3
Calculate the remaining thermal energy available to increase the temperature.
Qrem=207,000 J165,000 J=42,000 JQ_{\text{rem}} = 207,000\text{ J} - 165,000\text{ J} = 42,000\text{ J}.
After complete melting, excess energy goes into sensible heating.
4
Calculate the total mass of liquid water and the resulting temperature rise.
mtotal=0.50 kg (melted ice)+0.50 kg (initial water)=1.00 kgm_{\text{total}} = 0.50\text{ kg (melted ice)} + 0.50\text{ kg (initial water)} = 1.00\text{ kg}. ΔT=Qremmtotal×cw=42,000 J1.00 kg×4200 J kg1 K1=10C\Delta T = \frac{Q_{\text{rem}}}{m_{\text{total}} \times c_w} = \frac{42,000\text{ J}}{1.00\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1}} = 10^\circ\text{C}.
All water now absorbs energy to raise the temperature.

Anahtar Kavram

Phase changes occur at constant temperature (latent heat Q=mLQ = mL). Once the phase change is complete, additional thermal energy increases temperature as sensible heat (Q=mcΔTQ = mc\Delta T) using the total combined mass.
Soru 4407Soru

Two point charges of +2.0×106 C+2.0 \times 10^{-6}\text{ C} and +4.0×106 C+4.0 \times 10^{-6}\text{ C} are placed in a vacuum at a distance of 0.3 m0.3\text{ m} apart. Taking Coulomb's constant k=9.0×109 Nm2C2k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}, what is the magnitude of the electrostatic force exerted between them in Newtons?

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Cevap: 0.8

Cevap

The magnitude of the electrostatic force between the charges is 0.8 N.
According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them: F=kq1q2r2F = \frac{k q_1 q_2}{r^2}. Substituting q1=2.0×106 Cq_1 = 2.0 \times 10^{-6}\text{ C}, q2=4.0×106 Cq_2 = 4.0 \times 10^{-6}\text{ C}, r=0.3 mr = 0.3\text{ m}, and k=9.0×109 Nm2C2k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2} into the formula gives F=0.8 NF = 0.8\text{ N}.

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1
Identify known quantities from the problem statement
q1=2.0×106 Cq_1 = 2.0 \times 10^{-6}\text{ C}, q2=4.0×106 Cq_2 = 4.0 \times 10^{-6}\text{ C}, r=0.3 mr = 0.3\text{ m}, k=9.0×109 Nm2C2k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}
Clear identification of parameters is required for substitution into Coulomb's Law.
2
Apply Coulomb's Law formula
F=kq1q2r2F = \frac{k q_1 q_2}{r^2}
Coulomb's Law quantifies the electrostatic force between two stationary point charges.
3
Substitute values and perform arithmetic calculation
F=9.0×109×(2.0×106)×(4.0×106)0.09=0.0720.09=0.8 NF = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}
Squaring the separation distance 0.3 m0.3\text{ m} gives 0.09 m20.09\text{ m}^2, and evaluating the numerator gives 0.072 Nm20.072\text{ N}\cdot\text{m}^2.

Anahtar Kavram

Coulomb's Law
Soru 4408Soru

A metal rod of length 0.4 m0.4\text{ m} glides at a constant speed of 5.0 m s15.0\text{ m s}^{-1} to the right along horizontal parallel conducting rails placed in a uniform magnetic field of 0.5 T0.5\text{ T} directed vertically into the page. The rails are connected at their left end by a 2.0 Ω2.0\ \Omega resistor. What is the magnitude of the external force required to maintain the uniform speed of the rod, and in which direction does the induced magnetic force act on the rod?

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Cevap: 0.10 N0.10\text{ N}, directed to the left

Cevap

0.10 N0.10\text{ N}, directed to the left
The motional e.m.f. generated across the moving conductor is E=BLv=0.5×0.4×5.0=1.0 V\mathcal{E} = BLv = 0.5 \times 0.4 \times 5.0 = 1.0\text{ V}. The induced current is I=E/R=1.0/2.0=0.5 AI = \mathcal{E}/R = 1.0 / 2.0 = 0.5\text{ A}. The magnetic force resisting the rod's motion is FB=BIL=0.5×0.5×0.4=0.10 NF_B = BIL = 0.5 \times 0.5 \times 0.4 = 0.10\text{ N}. By Lenz's law, this magnetic force acts to the left, opposing the motion to the right. To maintain uniform speed, an equal external force of 0.10 N0.10\text{ N} must be applied.

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1
Calculate the induced electromotive force (e.m.f.) across the moving rod
E=BLv=0.5 T×0.4 m×5.0 m s1=1.0 V\mathcal{E} = B L v = 0.5\text{ T} \times 0.4\text{ m} \times 5.0\text{ m s}^{-1} = 1.0\text{ V}
Motional e.m.f. is produced when a conductor cuts magnetic flux lines at a perpendicular velocity.
2
Calculate the induced current flowing in the circuit
I=ER=1.0 V2.0 Ω=0.5 AI = \frac{\mathcal{E}}{R} = \frac{1.0\text{ V}}{2.0\ \Omega} = 0.5\text{ A}
Ohm's law relates the induced e.m.f. and total circuit resistance.
3
Determine the magnitude of the magnetic force acting on the current-carrying rod
FB=BIL=0.5 T×0.5 A×0.4 m=0.10 NF_B = B I L = 0.5\text{ T} \times 0.5\text{ A} \times 0.4\text{ m} = 0.10\text{ N}
A magnetic field exerts a force on a straight conductor carrying current.
4
Determine the direction of the magnetic force using Lenz's law and Newton's first law
The magnetic force acts to the left (opposing motion to the right). An external force of equal magnitude (0.10 N0.10\text{ N}) to the right is required to maintain constant speed.
Lenz's law states that induced effects always oppose the change causing them (motion to the right).

Anahtar Kavram

Motional Electromotive Force, Magnetic Force on a Conductor, and Lenz's Law
Soru 4409Soru

An inclined plane of length 5 m5\text{ m} is used to lift a load of 400 N400\text{ N} through a vertical height of 1 m1\text{ m}. If an effort of 100 N100\text{ N} is applied parallel to the incline, what is the efficiency of the machine?

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Cevap: 80%80\%

Cevap

The efficiency of the machine is 80%80\%.
The correct answer of 80%80\% is obtained by finding the velocity ratio (distance moved by effort divided by distance moved by load, 5/1=55 / 1 = 5) and the mechanical advantage (load divided by effort, 400/100=4400 / 100 = 4), then calculating efficiency as (MA/VR)×100%=(4/5)×100%=80%(\text{MA} / \text{VR}) \times 100\% = (4 / 5) \times 100\% = 80\%.

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1
Calculate the Velocity Ratio (VR) of the inclined plane
VR=Length of inclineHeight=5 m1 m=5\text{VR} = \frac{\text{Length of incline}}{\text{Height}} = \frac{5\text{ m}}{1\text{ m}} = 5
Velocity Ratio is the distance moved by the effort divided by the distance moved by the load.
2
Calculate the Mechanical Advantage (MA)
MA=LoadEffort=400 N100 N=4\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4
Mechanical Advantage measures how many times a machine multiplies the applied force.
3
Calculate the Efficiency
Efficiency=(MAVR)×100%=(45)×100%=80%\text{Efficiency} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4}{5}\right) \times 100\% = 80\%
Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Anahtar Kavram

Efficiency of Simple Machines (Inclined Plane)
Tahmini Süre:45s
Soru 4410Soru

A hypermetropic eye has its unassisted near point situated at 75 cm75\text{ cm} from the eye. Calculate the focal length, in cm\text{cm}, of the converging spectacle lens needed to enable the person to read a book comfortably at a distance of 25 cm25\text{ cm} from the eye.

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Cevap: 37.5

Cevap

The focal length of the required converging lens is 37.5 cm37.5\text{ cm}.
To correct hypermetropia, the spectacle lens must form a virtual image of an object located at the desired near point (u=+25 cmu = +25\text{ cm}) at the eye's actual, unassisted near point (v=75 cmv = -75\text{ cm}). Substituting u=+25 cmu = +25\text{ cm} and v=75 cmv = -75\text{ cm} into the thin lens formula 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} gives 1f=125175=275 cm1\frac{1}{f} = \frac{1}{25} - \frac{1}{75} = \frac{2}{75}\text{ cm}^{-1}. Taking the reciprocal yields f=37.5 cmf = 37.5\text{ cm}.

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1
Determine the object distance and image distance with sign conventions
u=+25 cmu = +25\text{ cm} and v=75 cmv = -75\text{ cm}.
The object is placed at the normal reading distance (25 cm25\text{ cm}), and the spectacle lens creates a virtual image on the same side of the lens at the person's near point (75 cm75\text{ cm}).
2
Set up the thin lens formula
1f=125175.\frac{1}{f} = \frac{1}{25} - \frac{1}{75}.
The thin lens formula relates the focal length to object and image distances.
3
Calculate the focal length ff
f = 37.5\text{ cm}.
Simplifying 3175=275\frac{3 - 1}{75} = \frac{2}{75} and taking the reciprocal yields f=752=37.5 cmf = \frac{75}{2} = 37.5\text{ cm}.

Anahtar Kavram

Correction of hypermetropia (farsightedness) using thin lens formula with virtual image sign convention
Soru 4411Soru

A variable force FF acts on a body moving along a straight horizontal path. The force increases linearly from 0 N0\text{ N} at position x=0 mx = 0\text{ m} to 20 N20\text{ N} at x=4 mx = 4\text{ m}, remains constant at 20 N20\text{ N} from x=4 mx = 4\text{ m} to x=7 mx = 7\text{ m}, and then decreases linearly back to 0 N0\text{ N} at x=10 mx = 10\text{ m}. What is the total work done by the force in Joules over the 10 m10\text{ m} displacement?

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Cevap: 130

Cevap

130 J
The work done by a variable force is equal to the total area under the force-displacement (FF-xx) graph. The region under the graph forms a trapezoid bounded by parallel sides of length 10 m10\text{ m} (total displacement) and 3 m3\text{ m} (constant force interval), with a height of 20 N20\text{ N}. Calculating the area yields W=12(10+3)×20=130 JW = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}.

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1
Relate work done to the force-displacement graph
Work done WW equals the total area under the FF-xx graph between x=0 mx = 0\text{ m} and x=10 mx = 10\text{ m}.
By definition, W=FdxW = \int F \, dx, which corresponds to the geometric area under the force-displacement curve.
2
Calculate the geometric area under each section of the graph
First section (00 to 4 m4\text{ m}): Triangle area = 12×4×20=40 J\frac{1}{2} \times 4 \times 20 = 40\text{ J}. Second section (44 to 7 m7\text{ m}): Rectangle area = (74)×20=60 J(7 - 4) \times 20 = 60\text{ J}. Third section (77 to 10 m10\text{ m}): Triangle area = 12×(107)×20=30 J\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}.
Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.
3
Sum the areas of all sections
Total Work W=40 J+60 J+30 J=130 JW = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}.
The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Anahtar Kavram

Work Done by a Variable Force (Area under Force-Displacement Graph)
Soru 4412Soru

Two marksmen, Kemi and Chidi, independently shoot at a target once. The probability that Kemi hits the target is 0.70.7, and the probability that Chidi hits the target is 0.60.6. What is the probability that at least one of them hits the target?

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Cevap: 0.88

Cevap

The probability that at least one marksman hits the target is 0.88.
Since the two shooting attempts are independent, the probability that both marksmen miss is (10.7)×(10.6)=0.3×0.4=0.12(1 - 0.7) \times (1 - 0.6) = 0.3 \times 0.4 = 0.12. Subtracting this probability from 1 yields 10.12=0.881 - 0.12 = 0.88, which represents the probability that at least one marksman hits the target.

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1
Find the probability of each event not occurring (missing the target).
P(Kemi misses)=0.3P(\text{Kemi misses}) = 0.3 and P(Chidi misses)=0.4P(\text{Chidi misses}) = 0.4
The sum of an event's probability and its complement is always 1.
2
Determine the probability of both events failing simultaneously.
P(both miss)=0.3×0.4=0.12P(\text{both miss}) = 0.3 \times 0.4 = 0.12
For independent events AA and BB, P(AB)=P(A)×P(B)P(A' \cap B') = P(A') \times P(B').
3
Use the complement rule to find the probability of at least one success.
P(at least one hits)=10.12=0.88P(\text{at least one hits}) = 1 - 0.12 = 0.88
The event 'at least one hits' is the exact complement of 'neither hits'.

Anahtar Kavram

Compound probability laws for independent events and complement of combined events
Soru 4413Soru

A uniform metal rod of length 0.50 m0.50\text{ m} and cross-sectional area 2.0×103 m22.0 \times 10^{-3}\text{ m}^2 has a thermal conductivity of 400 Wm1K1400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}. If a temperature difference of 50 K50\text{ K} is maintained between its ends, what is the rate of heat flow through the rod?

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Cevap: 80 W80\text{ W}

Cevap

The rate of heat flow through the rod is 80 W80\text{ W}.
The rate of heat transfer by conduction is calculated using Fourier's law, Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}. Substituting k=400 Wm1K1k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, A=2.0×103 m2A = 2.0 \times 10^{-3}\text{ m}^2, ΔT=50 K\Delta T = 50\text{ K}, and d=0.50 md = 0.50\text{ m} yields 80 W80\text{ W}.

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1
Identify the given parameters for thermal conduction.
Thermal conductivity k=400 Wm1K1k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, cross-sectional area A=2.0×103 m2A = 2.0 \times 10^{-3}\text{ m}^2, temperature difference ΔT=50 K\Delta T = 50\text{ K}, and length d=0.50 md = 0.50\text{ m}.
These parameters form the components of Fourier's law of thermal conduction.
2
Apply the conduction rate formula Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}.
\frac{Q}{t} = \frac{400 \times (2.0 \times 10^{-3}) \times 50}{0.50}
Heat transfer per unit time depends directly on conductivity, area, and temperature difference, and inversely on thickness/length.
3
Calculate the numerical value.
\frac{Q}{t} = \frac{40}{0.50} = 80\text{ W}
Dividing the product of the numerator terms (40 J/sm40\text{ J/s}\cdot\text{m}) by length (0.50 m0.50\text{ m}) yields the rate of energy transfer.

Anahtar Kavram

Thermal Conduction Rate (Fourier's Law)
Tahmini Süre:45s
Soru 4414Soru

A crate on a smooth surface is simultaneously pulled by two horizontal forces. One force of magnitude 9.0 N9.0\text{ N} acts towards the East, and another force of magnitude 12.0 N12.0\text{ N} acts towards the North. What is the magnitude of the resultant force acting on the crate in newtons?

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Cevap: 15

Cevap

The magnitude of the resultant force acting on the crate is 15.0 N15.0\text{ N}.
Because the two pulling forces are perpendicular, their resultant magnitude is calculated using vector addition via the Pythagorean theorem: R=9.02+12.02=15.0 NR = \sqrt{9.0^2 + 12.0^2} = 15.0\text{ N}.

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1
Determine the angle between the two given vectors
The forces are perpendicular (9090^\circ).
East and North cardinal directions are orthogonal to each other.
2
Compute the resultant magnitude using vector synthesis
R=9.02+12.02=81+144=225=15.0 NR = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ N}
For perpendicular force vectors, the resultant magnitude is the hypotenuse of the right triangle formed by the vector components.

Anahtar Kavram

Vector addition of perpendicular forces using the Pythagorean theorem
Tahmini Süre:1m 15s
Soru 4415Soru

Read the excerpt below carefully:

*During the tribunal hearing, the magistrate observed that the defendant's guilt was an open secret among the townspeople long before the trial commenced.*

Which figure of speech is expressed by the highlighted phrase, and what is its contextual meaning in the sentence?

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Cevap: An oxymoron indicating a fact that is officially unacknowledged yet known to almost everyone

Cevap

An oxymoron indicating a fact that is officially unacknowledged yet known to almost everyone
The phrase 'open secret' is an oxymoron because it combines two contradictory concepts: 'open' (publicly visible) and 'secret' (concealed). In English lexis and literary usage, this expression connotes a situation or piece of information that is officially kept quiet or unverified, yet is widely known and understood by the public.

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1
Identify the literary device used in the highlighted phrase 'open secret'.
The phrase places two contradictory terms side by side ('open', meaning public/unconcealed, and 'secret', meaning hidden/confidential). This juxtaposition of contradictory words forms an oxymoron.
Recognizing the structural relationship between opposing terms reveals the underlying figure of speech.
2
Analyze the contextual connotation of the phrase within the sentence.
In the context of the tribunal hearing, saying something is an 'open secret' means that while it is not formally announced or acknowledged, practically everyone in the community is already aware of it.
Figurative expressions must be interpreted according to their pragmatic contextual usage rather than literal word-for-word definitions.

Anahtar Kavram

Oxymoron and Figurative Connotation
Soru 4416Soru

A sample of gas enclosed in a rigid container exerts a pressure of 3.0×105 N m23.0 \times 10^5 \text{ N m}^{-2} with a density of 0.40 kg m30.40 \text{ kg m}^{-3}. If the gas is heated at constant volume until its pressure increases to 1.2×106 N m21.2 \times 10^6 \text{ N m}^{-2}, what is the final root-mean-square (r.m.s.) speed of the gas molecules?

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Cevap: 3000 m s13000 \text{ m s}^{-1}

Cevap

The final root-mean-square speed of the gas molecules is 3000 m s13000 \text{ m s}^{-1}.
Using the kinetic theory equation P=13ρvrms2P = \frac{1}{3}\rho v_{\text{rms}}^2, the initial speed is v1=3×3.0×1050.40=1500 m s1v_1 = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.40}} = 1500 \text{ m s}^{-1}. When heated at constant volume, density remains unchanged. The new pressure 1.2×106 N m21.2 \times 10^6 \text{ N m}^{-2} is 4 times the initial pressure, so the new speed is v2=4×v1=2×1500 m s1=3000 m s1v_2 = \sqrt{4} \times v_1 = 2 \times 1500 \text{ m s}^{-1} = 3000 \text{ m s}^{-1}.

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1
Express the relationship between pressure, density, and root-mean-square speed using kinetic theory.
P=13ρvrms2    vrms=3PρP = \frac{1}{3}\rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}
According to the kinetic theory of gases, the pressure exerted by a gas is related to its density and molecular r.m.s. speed.
2
Calculate the initial root-mean-square speed v1v_1 using initial pressure P1=3.0×105 N m2P_1 = 3.0 \times 10^5 \text{ N m}^{-2} and density ρ=0.40 kg m3\rho = 0.40 \text{ kg m}^{-3}.
v1=3×(3.0×105)0.40=9.0×1050.40=2.25×106=1500 m s1v_1 = \sqrt{\frac{3 \times (3.0 \times 10^5)}{0.40}} = \sqrt{\frac{9.0 \times 10^5}{0.40}} = \sqrt{2.25 \times 10^6} = 1500 \text{ m s}^{-1}
Establishes the baseline molecular speed prior to heating.
3
Determine the final root-mean-square speed v2v_2 after the pressure increases to P2=1.2×106 N m2P_2 = 1.2 \times 10^6 \text{ N m}^{-2} at constant volume.
v2=3×(1.2×106)0.40=3.6×1060.40=9.0×106=3000 m s1v_2 = \sqrt{\frac{3 \times (1.2 \times 10^6)}{0.40}} = \sqrt{\frac{3.6 \times 10^6}{0.40}} = \sqrt{9.0 \times 10^6} = 3000 \text{ m s}^{-1}
Since the volume is rigid, density ρ\rho remains constant while pressure increases due to heating.

Anahtar Kavram

Root-Mean-Square Speed and Pressure Relation in Kinetic Theory
Soru 4417Soru

When a strong bar magnet is dropped vertically through a long, hollow copper tube, its downward acceleration is equal to the acceleration due to gravity (gg) because copper is a non-magnetic material.

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Cevap: False

Cevap

The statement is false. The magnet falls with an acceleration less than gg because induced eddy currents in the copper tube create an upward magnetic force that opposes the motion.
The statement is false because the relative motion between the falling magnet and the conductive copper tube induces eddy currents. According to Lenz's law, these induced currents set up a magnetic field that opposes the falling magnet's motion, creating an upward retarding force that reduces the downward acceleration to a value less than gg.

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1
Identify the physical interactions as the magnet falls through the tube.
The falling magnet creates a changing magnetic flux through the surrounding copper tube.
Relative motion between a magnetic field source and a conductor produces a time-varying magnetic flux in the conductor.
2
Apply Faraday's law of electromagnetic induction.
Electromotive force (e.m.f.) and circular eddy currents are induced in the conductive copper walls.
A changing magnetic flux induces electric currents in any closed conductive path.
3
Apply Lenz's law to determine the magnetic effect of the induced eddy currents.
The induced eddy currents produce a magnetic field that opposes the downward motion of the falling magnet, generating an upward magnetic force (FmagF_{\text{mag}}).
Lenz's law dictates that an induced current always flows in a direction such that its magnetic field opposes the change causing it.
4
Analyze the net force and resulting acceleration.
The net downward force is Fnet=mgFmag<mgF_{\text{net}} = mg - F_{\text{mag}} < mg, so the downward acceleration a=gFmagm<ga = g - \frac{F_{\text{mag}}}{m} < g.
The presence of an upward magnetic force reduces the net downward acceleration below the free-fall value of gg.

Anahtar Kavram

Lenz's Law and Eddy Currents in Conductors
Tahmini Süre:1m 0s
Soru 4418Soru

A stretched string of length 0.50 m0.50\text{ m} fixed at both ends vibrates in its third harmonic mode at a frequency of 450 Hz450\text{ Hz}. What is the speed of the transverse wave along the string in m/s\text{m/s}?

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Cevap: 150

Cevap

The speed of the transverse wave along the string is 150 m/s150\text{ m/s}.
For a string fixed at both ends, standing wave modes produce harmonics given by fn=nv2Lf_n = \frac{n v}{2L}. Given n=3n = 3, L=0.50 mL = 0.50\text{ m}, and f3=450 Hzf_3 = 450\text{ Hz}, substituting these into the equation yields 450=3v2(0.50)=3v450 = \frac{3v}{2(0.50)} = 3v, leading to v=150 m/sv = 150\text{ m/s}.

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1
Identify the standing wave frequency equation for a string fixed at both ends.
The frequency of the nn-th harmonic is fn=nv2Lf_n = \frac{n v}{2L}, where nn is the harmonic number, vv is the wave speed, and LL is the string length.
Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths λn=2Ln\lambda_n = \frac{2L}{n}.
2
Substitute the given physical quantities into the harmonic equation.
450=3×v2×0.50450 = \frac{3 \times v}{2 \times 0.50}.
The question specifies L=0.50 mL = 0.50\text{ m}, third harmonic mode (n=3n = 3), and frequency f3=450 Hzf_3 = 450\text{ Hz}.
3
Solve for the wave speed vv.
v=150 m/sv = 150\text{ m/s}.
Simplifying 2×0.50=1.02 \times 0.50 = 1.0 gives 3v=4503v = 450, so v=4503=150 m/sv = \frac{450}{3} = 150\text{ m/s}.

Anahtar Kavram

Standing Waves and Harmonics in Vibrating Strings
Soru 4419Soru

A uniform metallic wire of resistance RR is stretched uniformly until its radius decreases by 20%20\%. The stretched wire is subsequently cut into two equal halves, which are then connected in parallel across a constant potential difference VV. What is the ratio of the total electrical power dissipated in this parallel combination to the power dissipated by the original unstretched wire under the same potential difference?

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Cevap: 1.64

Cevap

The ratio of the total power dissipated in the parallel combination to the original power is 1.64
When a wire of initial resistance RR is stretched so that its radius decreases by 20%20\%, its new radius is 0.8r0.8r. Because volume is conserved (A1L1=A2L2A_1 L_1 = A_2 L_2), reducing the cross-sectional area to 0.64A0.64A causes the length to increase to L/0.64L/0.64. Consequently, the resistance scales inversely with the fourth power of the radius: Rstretched=R/(0.8)4=R/0.4096=2.4414RR_{\text{stretched}} = R / (0.8)^4 = R / 0.4096 = 2.4414R. Cutting this wire into two equal pieces gives two resistors of 1.2207R1.2207R each. Connecting them in parallel yields an equivalent resistance Req=1.2207R/2=0.61035RR_{\text{eq}} = 1.2207R / 2 = 0.61035R. Power at constant voltage is P=V2/RP = V^2/R, so the new power is Pnew=V2/(0.61035R)=1.64(V2/R)=1.64PorigP_{\text{new}} = V^2 / (0.61035R) = 1.64 (V^2/R) = 1.64 P_{\text{orig}}.

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1
Determine the new resistance of the wire after stretching
Rstretched=R(0.8)4=R0.40962.4414RR_{\text{stretched}} = \frac{R}{(0.8)^4} = \frac{R}{0.4096} \approx 2.4414 R
Since mass and density remain constant, volume Vvol=ALV_{\text{vol}} = A \cdot L is conserved. Decreasing radius to r2=0.8r1r_2 = 0.8 r_1 reduces area to A2=0.64A1A_2 = 0.64 A_1 and increases length to L2=L1/0.64L_2 = L_1 / 0.64. Resistance R=ρL/A1/r4R = \rho L / A \propto 1/r^4.
2
Calculate the equivalent resistance of the two equal halves connected in parallel
Req=14Rstretched=2.4414R40.61035RR_{\text{eq}} = \frac{1}{4} R_{\text{stretched}} = \frac{2.4414 R}{4} \approx 0.61035 R
Cutting the stretched wire in half gives two pieces each of resistance Rhalf=Rstretched/2R_{\text{half}} = R_{\text{stretched}} / 2. Connecting two identical resistors in parallel yields an equivalent resistance Req=Rhalf/2=Rstretched/4R_{\text{eq}} = R_{\text{half}} / 2 = R_{\text{stretched}} / 4.
3
Calculate the ratio of power dissipated across a constant potential difference V
PnewPorig=V2/ReqV2/R=RReq=10.610351.64\frac{P_{\text{new}}}{P_{\text{orig}}} = \frac{V^2 / R_{\text{eq}}}{V^2 / R} = \frac{R}{R_{\text{eq}}} = \frac{1}{0.61035} \approx 1.64
Electrical power dissipated at constant voltage is given by P=V2/RP = V^2 / R, which means power is inversely proportional to equivalent resistance.

Anahtar Kavram

Dependence of electrical resistance on conductor geometry under volume conservation, and power dissipation in parallel circuits.
Soru 4420Soru

A thermometer calibrated on an arbitrary scale XX registers a lower fixed point of 10X-10^\circ\text{X} and an upper fixed point of 110X110^\circ\text{X}. What is the true temperature in degrees Celsius (C^\circ\text{C}) when this thermometer reads 20X20^\circ\text{X}?

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Cevap: 25

Cevap

The true temperature on the Celsius scale is 25C25^\circ\text{C}.
Using the relation XLFPXUFPXLFPX=θ100\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}, substituting X=20X = 20, LFPX=10\text{LFP}_X = -10, and UFPX=110\text{UFP}_X = 110 gives 20(10)110(10)=30120=0.25\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25. Multiplying 0.250.25 by 100100 gives 25C25^\circ\text{C}.

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1
Set up the linear relationship between the arbitrary temperature scale XX and the Celsius scale
XLFPXUFPXLFPX=θLFPCUFPCLFPC\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}
Thermometric properties vary linearly with temperature between fixed points.
2
Substitute the given numerical values into the formula
20(10)110(10)=θ01000    30120=θ100\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}
The lower fixed point on scale XX is 10X-10^\circ\text{X} and the upper fixed point is 110X110^\circ\text{X}.
3
Solve for the unknown temperature θ\theta in degrees Celsius
θ=30120×100=25C\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}
Simplifying the fraction 30120\frac{30}{120} yields 14\frac{1}{4}, and 14×100=25\frac{1}{4} \times 100 = 25.

Anahtar Kavram

Linear interpolation and conversion between thermometric temperature scales
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