Tüm alıştırma soruları

13931 soru

Soru 7981Soru

A binary operation \star defined on the set of real numbers R\mathbb{R} is given by xy=x+yxy+1x \star y = \frac{x + y}{x - y + 1} for xy1x - y \neq -1. If 5p=35 \star p = 3, what is the value of pp?

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Cevap: 3.25

Cevap

The value of pp is 3.253.25 (or 134\frac{13}{4}).
Applying the operation rule xy=x+yxy+1x \star y = \frac{x + y}{x - y + 1} with x=5x = 5 and y=py = p gives 5+p6p=3\frac{5 + p}{6 - p} = 3. Cross-multiplying gives 5+p=183p5 + p = 18 - 3p, which simplifies to 4p=134p = 13 or p=3.25p = 3.25.

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1
Substitute x=5x = 5 and y=py = p into the given binary operation definition
5+p5p+1=3\frac{5 + p}{5 - p + 1} = 3
This sets up the equation for the given condition 5p=35 \star p = 3.
2
Simplify the denominator in the algebraic fraction
5+p6p=3\frac{5 + p}{6 - p} = 3
Combining the constants 5+1=65 + 1 = 6 simplifies the denominator.
3
Multiply both sides by (6p)(6 - p) and expand
5+p=183p5 + p = 18 - 3p
Eliminating the denominator allows linear terms in pp to be collected.
4
Rearrange terms to solve for pp
4p=13    p=3.254p = 13 \implies p = 3.25
Adding 3p3p to both sides and subtracting 55 gives 4p=134p = 13.

Anahtar Kavram

Solving linear equations derived from non-commutative binary operations
Soru 7982Soru

Which of the following combinations consists entirely of derived physical quantities?

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Cevap: Volume, momentum, and electrical potential difference

Cevap

The combination comprising volume, momentum, and electrical potential difference consists entirely of derived physical quantities.
The combination containing volume, momentum, and electrical potential difference consists entirely of derived physical quantities because volume depends on length (L3L^3), momentum depends on mass, length, and time (MLT1M\cdot L\cdot T^{-1}), and electrical potential difference depends on mass, length, time, and electric current (ML2T3I1M\cdot L^2\cdot T^{-3}\cdot I^{-1}). None of these three are base/fundamental quantities.

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1
Identify the seven fundamental physical quantities in SI units.
The seven fundamental quantities are mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.
Fundamental quantities serve as the basic foundation from which all other physical quantities are defined.
2
Classify each physical quantity present in the given combinations.
Volume (m3m^3), momentum (kgm/skg\cdot m/s), and potential difference (VV or kgm2/(As3)kg\cdot m^2/(A\cdot s^3)) are all derived from fundamental units. Other combinations contain fundamental quantities such as length, mass, time, temperature, or electric current.
Derived quantities are physical quantities defined by mathematical combinations of fundamental quantities.
3
Select the option where every listed quantity is a derived quantity.
The group containing volume, momentum, and electrical potential difference is the only set where all items are derived quantities.
It fulfills the requirement of containing exclusively derived physical quantities.

Anahtar Kavram

Fundamental quantities are independent quantities defined by international standard protocols, whereas derived quantities are formed by mathematical combinations of fundamental quantities.
Soru 7983Soru

An electric immersion heater with a resistance of 20Ω20\,\Omega carries a steady current of 3A3\,\text{A}. How much electrical energy is dissipated as heat by the heater in 10seconds10\,\text{seconds}?

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Cevap: 1800J1800\,\text{J}

Cevap

The total electrical energy dissipated as heat by the heater is 1800J1800\,\text{J}.
According to Joule's law of heating, the electrical energy EE converted into heat energy in a resistor is given by E=I2RtE = I^2 R t. Substituting I=3AI = 3\,\text{A}, R=20ΩR = 20\,\Omega, and t=10st = 10\,\text{s} yields E=(3)2×20×10=9×200=1800JE = (3)^2 \times 20 \times 10 = 9 \times 200 = 1800\,\text{J}.

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1
Identify the given physical quantities
Resistance R=20ΩR = 20\,\Omega, Current I=3AI = 3\,\text{A}, Time t=10st = 10\,\text{s}
Recognize the required values needed to compute heat energy dissipated in a resistor.
2
Select the appropriate formula for heat energy
E=I2RtE = I^2 R t
Joule's Law of Heating relates electrical energy, current, resistance, and time.
3
Substitute the known values into the equation and calculate
E=(3)2×20×10=9×20×10=1800JE = (3)^2 \times 20 \times 10 = 9 \times 20 \times 10 = 1800\,\text{J}
Perform arithmetic multiplication to find the final heat energy in Joules.

Anahtar Kavram

Joule's Law of Heating
Soru 7984Soru

A block of mass 0.5 kg0.5\text{ kg} absorbs 4500 J4500\text{ J} of heat energy as its temperature rises from 25C25^\circ\text{C} to 45C45^\circ\text{C}. What is the heat capacity of the block?

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Cevap: 225 J K1225\text{ J K}^{-1}

Cevap

225 J K1225\text{ J K}^{-1}
Heat capacity (CC) is defined as thermal energy absorbed divided by temperature change: C=QΔT=4500 J45 K25 K=225 J K1C = \frac{Q}{\Delta T} = \frac{4500\text{ J}}{45\text{ K} - 25\text{ K}} = 225\text{ J K}^{-1}.

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1
Calculate the temperature change (ΔT\Delta T).
ΔT=45C25C=20C=20 K\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20^\circ\text{C} = 20\text{ K}
Heat capacity depends on the temperature change in Kelvin or degrees Celsius.
2
Apply the heat capacity formula C=QΔTC = \frac{Q}{\Delta T}.
C=4500 J20 K=225 J K1C = \frac{4500\text{ J}}{20\text{ K}} = 225\text{ J K}^{-1}
Heat capacity (CC) measures the heat required to raise the temperature of the entire body by 1 K1\text{ K}, regardless of mass.

Anahtar Kavram

Heat capacity (CC) represents the energy required to change an entire object's temperature by one kelvin (C=QΔTC = \frac{Q}{\Delta T}), whereas specific heat capacity (cc) is per unit mass (c=QmΔTc = \frac{Q}{m\Delta T}).
Tahmini Süre:1m 0s
Soru 7985Soru

For an object of constant mass moving at constant speed along a complete circular path, the net vector impulse imparted to the object over one full revolution is zero, even though a continuous centripetal force acts on the object throughout the motion.

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Cevap: True

Cevap

The statement is true because the impulse-momentum theorem states that net impulse equals the change in momentum (J=Δp\vec{J} = \Delta \vec{p}). After one full revolution, the object's initial and final velocity vectors are identical, resulting in zero change in momentum.
The statement is correct because linear momentum is a vector quantity. Over a complete circular revolution, the initial and final velocity vectors are identical in both magnitude and direction, making the change in momentum—and therefore the net vector impulse—equal to zero.

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1
Recall the vector definition of impulse and the impulse-momentum theorem.
The net impulse J\vec{J} acting on a body equals its change in linear momentum: J=Δp=pfpi=mvfmvi\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i.
Impulse is a vector quantity dependent on the initial and final states of momentum over the time interval.
2
Evaluate the velocity vector of the object after one complete circular revolution at constant speed.
Since the speed is constant and the trajectory completes a closed loop, the final velocity vector vf\vec{v}_f has the exact same magnitude and direction as the initial velocity vector vi\vec{v}_i.
A complete revolution returns the object to its starting point with its velocity pointing in the initial direction.
3
Calculate the change in momentum Δp\Delta \vec{p}.
Δp=m(vfvi)=m(0)=0\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(0) = \vec{0}. Thus, net impulse J=0\vec{J} = \vec{0}.
Subtracting identical vectors yields zero vector magnitude.

Anahtar Kavram

Impulse-Momentum Theorem and Vector Nature of Linear Momentum
Tahmini Süre:1m 30s
Soru 7986Soru

A battery with an electromotive force (e.m.f.) of 12.0 V12.0\text{ V} and an internal resistance of 1.0 Ω1.0\text{ }\Omega is connected across a series combination of a 2.0 Ω2.0\text{ }\Omega resistor and a 3.0 Ω3.0\text{ }\Omega resistor. What is the potential difference across the terminals of the battery?

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Cevap: 10.0 V10.0\text{ V}

Cevap

10.0 V10.0\text{ V}
The total resistance of the circuit is the sum of the external load resistors and the internal resistance (2.0+3.0+1.0=6.0 Ω2.0 + 3.0 + 1.0 = 6.0\text{ }\Omega). The total current is I=12.0 V6.0 Ω=2.0 AI = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}. The potential difference across the terminals of the battery is the voltage across the external load V=I×Rext=2.0 A×5.0 Ω=10.0 VV = I \times R_{\text{ext}} = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V} (or EIr=12.02.0=10.0 VE - Ir = 12.0 - 2.0 = 10.0\text{ V}).

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1
Calculate total external resistance in series
Rext=2.0 Ω+3.0 Ω=5.0 ΩR_{\text{ext}} = 2.0\text{ }\Omega + 3.0\text{ }\Omega = 5.0\text{ }\Omega
Resistors in series add directly to give the total external load resistance.
2
Calculate total circuit resistance including internal resistance
Rtotal=Rext+r=5.0 Ω+1.0 Ω=6.0 ΩR_{\text{total}} = R_{\text{ext}} + r = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega
The cell's internal resistance is in series with the external circuit.
3
Determine the total current drawn from the battery
I=ERtotal=12.0 V6.0 Ω=2.0 AI = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}
Applying Ohm's law to the entire circuit.
4
Calculate terminal potential difference across the battery
V=EIr=12.0 V(2.0 A×1.0 Ω)=10.0 VV = E - Ir = 12.0\text{ V} - (2.0\text{ A} \times 1.0\text{ }\Omega) = 10.0\text{ V}
Terminal voltage is the e.m.f. minus the lost volts across internal resistance (or equivalently V=I×RextV = I \times R_{\text{ext}}).

Anahtar Kavram

Terminal potential difference vs. electromotive force (e.m.f.) and lost volts
Tahmini Süre:1m 30s
Soru 7987Soru

An electric crane lifts a load of 250 kg250\text{ kg} vertically upwards through a height of 12 m12\text{ m} in 10 s10\text{ s} at a constant speed. Taking the acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}, what is the useful output power of the crane in watts?

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Cevap: 3000

Cevap

The useful output power of the crane is 3000 W3000\text{ W}.
The work done in lifting the load vertically is equal to the gravitational potential energy gained, W=mgh=250×10×12=30,000 JW = mgh = 250 \times 10 \times 12 = 30,000\text{ J}. Power is the rate of doing work, so P=Wt=30,00010=3000 WP = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}.

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1
Identify the given values
Mass m=250 kgm = 250\text{ kg}, height h=12 mh = 12\text{ m}, time t=10 st = 10\text{ s}, acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}.
Extract values needed for work and power calculations.
2
Calculate the work done in lifting the load
W=mgh=250 kg×10 m s2×12 m=30,000 JW = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}.
The work done against gravity equals the gain in gravitational potential energy.
3
Calculate the power output
P=Wt=30,000 J10 s=3000 WP = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}.
Power is defined as the rate at which work is done (P=WtP = \frac{W}{t}).

Anahtar Kavram

Power as the rate of doing work against gravity
Soru 7988Soru

Given the matrices A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} and B=(1421)B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}, what is the product matrix ABAB?

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Cevap: (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

Cevap

The matrix product ABAB is (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.
To find ABAB, each element cijc_{ij} is formed by taking the dot product of row ii of matrix AA and column jj of matrix BB. Performing these steps gives c11=2(1)+1(2)=4c_{11} = 2(1) + 1(2) = 4, c12=2(4)+1(1)=7c_{12} = 2(4) + 1(-1) = 7, c21=0(1)+3(2)=6c_{21} = 0(1) + 3(2) = 6, and c22=0(4)+3(1)=3c_{22} = 0(4) + 3(-1) = -3, yielding (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.

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1
Multiply the first row of AA by the first column of BB to find element (1,1)(1,1).
2(1)+1(2)=2+2=42(1) + 1(2) = 2 + 2 = 4
Matrix multiplication rule requires taking the dot product of rows from the first matrix and columns from the second matrix.
2
Multiply the first row of AA by the second column of BB to find element (1,2)(1,2).
2(4)+1(1)=81=72(4) + 1(-1) = 8 - 1 = 7
Evaluates the element in row 1, column 2 of the resulting matrix.
3
Multiply the second row of AA by the first column of BB to find element (2,1)(2,1).
0(1)+3(2)=0+6=60(1) + 3(2) = 0 + 6 = 6
Evaluates the element in row 2, column 1 of the resulting matrix.
4
Multiply the second row of AA by the second column of BB to find element (2,2)(2,2).
0(4)+3(1)=03=30(4) + 3(-1) = 0 - 3 = -3
Evaluates the element in row 2, column 2 of the resulting matrix.

Anahtar Kavram

Matrix Multiplication
Tahmini Süre:1m 0s
Soru 7989Soru

A concave mirror forms a real image that is twice the size of an object. When the object is shifted 10 cm10\text{ cm} closer to the mirror, a virtual image of the same magnification is produced. What is the focal length of the mirror?

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Cevap: 10 cm10\text{ cm}

Cevap

The focal length of the concave mirror is 10 cm10\text{ cm}.
For a concave mirror forming a real image of magnification 22, v1=2u1v_1 = 2u_1, yielding u1=1.5fu_1 = 1.5f. When the object is moved 10 cm10\text{ cm} closer, a virtual image of magnification 22 is formed, so v2=2u2v_2 = -2u_2, yielding u2=0.5fu_2 = 0.5f. Subtracting the two object positions (1.5f0.5f=10 cm1.5f - 0.5f = 10\text{ cm}) directly gives f=10 cmf = 10\text{ cm}.

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1
Set up the mirror equation for the first case (real image).
u1=32fu_1 = \frac{3}{2}f
For a real inverted image with magnification m=2m = 2, v1=+2u1v_1 = +2u_1. Substituting into 1f=1u1+1v1\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1} gives 1f=1u1+12u1=32u1\frac{1}{f} = \frac{1}{u_1} + \frac{1}{2u_1} = \frac{3}{2u_1}.
2
Set up the mirror equation for the second case (virtual image).
u2=12fu_2 = \frac{1}{2}f
For a virtual erect image with magnification m=2m = 2, sign convention dictates v2=2u2v_2 = -2u_2. Substituting into 1f=1u2+1v2\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2} gives 1f=1u212u2=12u2\frac{1}{f} = \frac{1}{u_2} - \frac{1}{2u_2} = \frac{1}{2u_2}.
3
Use the given displacement between the two object positions to solve for ff.
f=10 cmf = 10\text{ cm}
The object is moved 10 cm10\text{ cm} closer, so u1u2=10 cmu_1 - u_2 = 10\text{ cm}. Substituting the expressions yields 32f12f=10 cm    f=10 cm\frac{3}{2}f - \frac{1}{2}f = 10\text{ cm} \implies f = 10\text{ cm}.

Anahtar Kavram

Mirror Formula and Sign Convention for Spherical Mirrors
Tahmini Süre:2m 0s
Soru 7990Soru

The volume flow rate QQ of a viscous liquid through a pipe depends on the radius rr of the pipe, the coefficient of viscosity η\eta, and the pressure gradient ΔPL\frac{\Delta P}{L} according to the dimensional equation Q=krxηy(ΔPL)zQ = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z, where kk is a dimensionless constant. What is the value of the exponent xx?

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Cevap: 4

Cevap

The value of the exponent xx is 4.
Applying the principle of dimensional homogeneity, the dimensions of volume flow rate [Q]=L3T1[Q] = L^3 T^{-1} are equated to [r]x[η]y[ΔPL]z=Lx(ML1T1)y(ML2T2)z[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z. Equating powers yields y+z=0y + z = 0 for mass, y2z=1-y - 2z = -1 for time, and xy2z=3x - y - 2z = 3 for length. Solving these simultaneous equations gives z=1z = 1, y=1y = -1, and x=4x = 4.

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1
Identify the base dimensions of each physical quantity in the given equation.
Flow rate [Q]=M0L3T1[Q] = M^0 L^3 T^{-1}, radius [r]=L[r] = L, viscosity [η]=ML1T1[\eta] = M L^{-1} T^{-1}, and pressure gradient [ΔPL]=ML2T2\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}.
Expressing each quantity in terms of fundamental dimensions (MM, LL, TT) is necessary for dimensional analysis.
2
Substitute dimensions into the power-law equation and collect powers of base dimensions.
M0L3T1=My+zLxy2zTy2zM^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}.
The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.
3
Set up and solve linear equations for the exponents xx, yy, and zz.
Solving y+z=0y + z = 0, y2z=1-y - 2z = -1, and xy2z=3x - y - 2z = 3 yields z=1z = 1, y=1y = -1, and x=4x = 4.
Equating powers of MM, TT, and LL allows step-by-step determination of each unknown exponent.

Anahtar Kavram

Dimensions of Physical Quantities and Dimensional Analysis
Soru 7991Soru

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency (ff) on the right, where vv is the speed of sound in air, LL is the physical length of the pipe or string, ee is the end correction per open end, TT is tension, and μ\mu is linear mass density.

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Öğeler

Fundamental mode of a pipe closed at one end, taking into account end correction
Fundamental mode of a uniform stretched string fixed at both ends
Fundamental mode of a pipe open at both ends, taking into account end corrections at both open ends
First overtone of a pipe closed at one end, neglecting end correction

Eşleşmeler

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Cevap

The fundamental mode of a pipe closed at one end with end correction matches f=v4(L+e)f = \frac{v}{4(L + e)}; the fundamental mode of a stretched string matches f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}; the fundamental mode of a pipe open at both ends with end correction at both ends matches f=v2(L+2e)f = \frac{v}{2(L + 2e)}; and the first overtone of a closed pipe without end correction matches f=3v4Lf = \frac{3v}{4L}.
Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency f=v4(L+e)f = \frac{v}{4(L+e)} for one open end, open pipes produce f=v2(L+2e)f = \frac{v}{2(L+2e)} for two open ends, stretched strings depend on tension and mass per unit length as f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, and the first overtone of a closed pipe is its third harmonic f=3v4Lf = \frac{3v}{4L}.

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1
Analyze boundary conditions and effective acoustic length for closed and open pipes.
A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction ee to its physical length LL (Leff=L+eL_{\text{eff}} = L + e). An open pipe has two open ends, giving an effective length Leff=L+2eL_{\text{eff}} = L + 2e.
Air displacement antinodes occur slightly outside open pipe boundaries by a distance ee per open end.
2
Derive the frequency formula for the fundamental mode of a closed pipe with end correction.
For the fundamental mode, L+e=λ4    λ=4(L+e)L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e). Frequency f=vλ=v4(L+e)f = \frac{v}{\lambda} = \frac{v}{4(L + e)}.
The distance between a node and an adjacent antinode is one-quarter of a wavelength.
3
Derive the fundamental frequency for a stretched string fixed at both ends.
L=λ2    λ=2LL = \frac{\lambda}{2} \implies \lambda = 2L. Using wave velocity v=Tμv = \sqrt{\frac{T}{\mu}}, f=v2L=12LTμf = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}.
Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.
4
Derive the fundamental frequency of an open pipe considering both end corrections.
L+2e=λ2    λ=2(L+2e)L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e), so f=v2(L+2e)f = \frac{v}{2(L + 2e)}.
Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.
5
Determine the first overtone frequency for a closed pipe without end correction.
The first overtone is the third harmonic (n=3n = 3), so L=3λ4    λ=4L3L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}, which gives f=3v4Lf = \frac{3v}{4L}.
Closed pipes support only odd integer multiples of the fundamental frequency.

Anahtar Kavram

Standing Waves and Resonance in Air Columns and Strings
Soru 7992Soru

A ticker-tape timer connected to a 50 Hz50\text{ Hz} alternating current mains supply is used to measure the time interval of a laboratory cart moving down an inclined plane. On the printed tape, 1010 distinct tick spaces are recorded between the initial and final position. A electronic stopwatch running concurrently has a positive zero error of +0.04 s+0.04\text{ s}. What is the true elapsed time interval represented by the ticker tape after properly correcting for the stopwatch zero error?

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Cevap: 0.16 s0.16\text{ s}

Cevap

The true elapsed time interval is 0.16 s0.16\text{ s}.
The ticker-tape frequency of 50 Hz50\text{ Hz} gives a time interval of 0.02 s0.02\text{ s} per space. Ten tick spaces correspond to an uncorrected time of 0.20 s0.20\text{ s}. To correct for a positive zero error of +0.04 s+0.04\text{ s}, the zero error must be subtracted from the uncorrected reading, yielding a true elapsed time of 0.16 s0.16\text{ s}.

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1
Calculate the period of a single tick interval
T=1f=150 Hz=0.02 sT = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}
Frequency ff is the inverse of period TT for a ticker-tape timer.
2
Determine the uncorrected total elapsed time from tick spaces
tuncorrected=10×0.02 s=0.20 st_{\text{uncorrected}} = 10 \times 0.02\text{ s} = 0.20\text{ s}
Elapsed time equals the number of tick spaces multiplied by the period of one space.
3
Apply the zero error correction
ttrue=tuncorrectedzero error=0.20 s0.04 s=0.16 st_{\text{true}} = t_{\text{uncorrected}} - \text{zero error} = 0.20\text{ s} - 0.04\text{ s} = 0.16\text{ s}
True Reading = Observed Reading - (Zero Error).

Anahtar Kavram

Measurement of time using ticker-tape timers and instrument zero error correction
Tahmini Süre:1m 30s
Soru 7993Soru

A hiker starts at camp CC and walks 5 km5\text{ km} due East to checkpoint AA. From checkpoint AA, the hiker then walks 3 km3\text{ km} on a bearing of 150150^\circ to reach checkpoint BB. What is the direct distance from camp CC to checkpoint BB?

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Cevap: 7 km7\text{ km}

Cevap

7 km7\text{ km}
The interior angle at checkpoint AA is calculated using three-figure bearings as 270150=120270^\circ - 150^\circ = 120^\circ. Applying the Cosine Rule CB2=52+322(5)(3)cos(120)CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ) yields 25+9+15=4925 + 9 + 15 = 49. Taking the square root gives the direct distance of 7 km7\text{ km}.

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1
Determine the interior angle CAB\angle CAB at checkpoint AA
CAB=120\angle CAB = 120^\circ
Due East corresponds to a bearing of 090090^\circ. Coming into AA from CC means line ACAC points West (270270^\circ). The bearing of BB from AA is 150150^\circ. The interior angle between vector ACAC pointing West (270270^\circ) and vector ABAB on bearing 150150^\circ is 270150=120270^\circ - 150^\circ = 120^\circ.
2
Apply the Cosine Rule to find length CBCB
CB2=CA2+AB22(CA)(AB)cos(CAB)CB^2 = CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB)
We have two sides (CA=5 kmCA = 5\text{ km}, AB=3 kmAB = 3\text{ km}) and the included angle (CAB=120\angle CAB = 120^\circ).
3
Substitute the known values into the Cosine Rule formula
CB2=52+322(5)(3)cos(120)=25+930(0.5)=34+15=49CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ) = 25 + 9 - 30(-0.5) = 34 + 15 = 49
Since cos(120)=cos(60)=0.5\cos(120^\circ) = -\cos(60^\circ) = -0.5, the negative sign inside the cosine product cancels with the subtraction sign in the formula.
4
Take the positive square root to find the distance
CB=49=7 kmCB = \sqrt{49} = 7\text{ km}
Distance must be positive.

Anahtar Kavram

Cosine Rule for non-right triangles in bearings problems
Soru 7994Soru

Determine whether the following statement is true or false: When a bar magnet is placed in the Earth's magnetic field with its north pole pointing towards geographic North, the neutral points formed by the mutual cancellation of the magnet's field and the Earth's horizontal field lie along the magnet's equatorial line (broadside-on position).

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Cevap: True

Cevap

True. When a bar magnet has its north pole pointing North, the magnetic field along its equatorial (broadside-on) line points South, opposing the Earth's horizontal magnetic field to create neutral points.
The statement is correct because neutral points are formed at positions where the external magnetic field of a magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field component (BHB_H). When the north pole of a bar magnet points North, its field along the broadside-on (equatorial) axis is directed South (opposite to BHB_H), leading to field cancellation and the creation of two neutral points on that axis.

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1
Identify the direction of the Earth's horizontal magnetic field component (BHB_H).
Earth's horizontal field (BHB_H) acts along the magnetic meridian from magnetic South to magnetic North.
Neutral points can only occur where two magnetic fields act in opposite directions and have equal magnitudes.
2
Determine the direction of the bar magnet's magnetic field on its broadside-on (equatorial) axis.
Outside the magnet, magnetic flux lines travel from North to South. Along the broadside-on line, the magnet's field points southward.
Field lines curve from the north pole back into the south pole externally.
3
Compare the magnetic field directions on the broadside-on axis.
The southward field of the magnet opposes the northward Earth field (BHB_H). At points where their magnitudes match, the net magnetic field becomes zero.
Vector cancellation requires antiparallel vectors of equal magnitude.

Anahtar Kavram

Neutral Points of a Bar Magnet in Earth's Magnetic Field
Soru 7995Soru

A student records the period of oscillation of a simple pendulum by measuring the time taken for 20 complete swings as 40.0 s40.0\text{ s}. If the absolute error in this time measurement is ±0.8 s\pm 0.8\text{ s}, what is the percentage error in the measured time?

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Cevap: 2.0%2.0\%

Cevap

2.0%2.0\%
The correct answer is 2.0%2.0\%. Percentage error is defined as the absolute error divided by the measured value, expressed as a percentage: 0.8 s40.0 s×100%=2.0%\frac{0.8\text{ s}}{40.0\text{ s}} \times 100\% = 2.0\%.

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1
Identify the given measurement and absolute error.
Measured time t=40.0 st = 40.0\text{ s} and absolute uncertainty Δt=0.8 s\Delta t = 0.8\text{ s}.
These quantities are needed to determine the relative error of the time measurement.
2
Apply the percentage error formula: Percentage Error=(Δtt)×100%\text{Percentage Error} = \left(\frac{\Delta t}{t}\right) \times 100\%.
Percentage Error=(0.8 s40.0 s)×100%=0.02×100%=2.0%\text{Percentage Error} = \left(\frac{0.8\text{ s}}{40.0\text{ s}}\right) \times 100\% = 0.02 \times 100\% = 2.0\%.
Multiplying the fractional error by 100 converts it into a percentage representation.

Anahtar Kavram

Percentage Error in Physical Measurements
Tahmini Süre:45s
Soru 7996Soru

A body of mass 2.5 kg2.5\text{ kg} moving at a speed of 4.0 m s14.0\text{ m s}^{-1} along a straight horizontal path is brought to rest in 2.0 s2.0\text{ s} by a constant retarding force. What is the magnitude of this retarding force in newtons?

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Cevap: 5

Cevap

The magnitude of the retarding force is 5.0 N5.0\text{ N}.
According to Newton's second law of motion, net force is equal to the rate of change of momentum (F=ΔpΔt=m(vu)tF = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}). Substituting m=2.5 kgm = 2.5\text{ kg}, u=4.0 m s1u = 4.0\text{ m s}^{-1}, v=0 m s1v = 0\text{ m s}^{-1}, and t=2.0 st = 2.0\text{ s} gives F=2.5×(04.0)2.0=5.0 NF = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}. The magnitude of this force is 5.0 N5.0\text{ N}.

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1
Determine the initial momentum and final momentum of the body.
Initial momentum pi=2.5×4.0=10.0 kg m s1p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}, and final momentum pf=0 kg m s1p_f = 0\text{ kg m s}^{-1}.
Linear momentum is defined as the product of mass and velocity (p=mvp = mv).
2
Calculate the magnitude of the force applied using the impulse-momentum relationship F=ΔpΔtF = \frac{\Delta p}{\Delta t}.
Magnitude of force F=010.02.0=5.0 NF = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}.
Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Anahtar Kavram

Newton's Second Law and Impulse-Momentum Relationship
Tahmini Süre:45s
Soru 7997Soru

A water wave traveling in deep water has a wavelength of 0.80 m0.80\text{ m} and a speed of 2.4 m/s2.4\text{ m/s}. Upon entering a shallow region, its speed drops to 1.8 m/s1.8\text{ m/s}. What are the frequency and wavelength of the wave in the shallow region?

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Cevap: Frequency = 3.0 Hz3.0\text{ Hz}, Wavelength = 0.60 m0.60\text{ m}

Cevap

Frequency = 3.0 Hz3.0\text{ Hz}, Wavelength = 0.60 m0.60\text{ m}
When a wave passes from deep to shallow water (refraction), its frequency remains unchanged because frequency is fixed by the source. Using v=fλv = f\lambda, the initial frequency is f=2.40.80=3.0 Hzf = \frac{2.4}{0.80} = 3.0\text{ Hz}. In shallow water, the new wavelength is λ=1.83.0=0.60 m\lambda' = \frac{1.8}{3.0} = 0.60\text{ m}.

Adım Adım Çözüm

1
Calculate the frequency of the wave in deep water using the wave equation v=fλv = f \lambda.
f=vλ=2.4 m/s0.80 m=3.0 Hzf = \frac{v}{\lambda} = \frac{2.4\text{ m/s}}{0.80\text{ m}} = 3.0\text{ Hz}.
The frequency depends on the wave source and can be determined from the given initial speed and wavelength.
2
Apply the boundary condition for wave refraction.
The frequency in shallow water remains f=3.0 Hzf = 3.0\text{ Hz}.
When a wave travels from one medium to another, its frequency remains constant.
3
Calculate the new wavelength in shallow water using λ=vf\lambda' = \frac{v'}{f}.
λ=1.8 m/s3.0 Hz=0.60 m\lambda' = \frac{1.8\text{ m/s}}{3.0\text{ Hz}} = 0.60\text{ m}.
The wavelength changes proportionally with speed when frequency is constant.

Anahtar Kavram

Constancy of wave frequency during refraction across medium boundaries
Soru 7998Soru

Match each of the physical quantities given on the left with its corresponding fundamental SI status or base unit resolution on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Luminous intensity
Linear momentum
Thermodynamic temperature
Specific heat capacity

Eşleşmeler

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Cevap

Luminous intensity matches with Fundamental quantity measured in candela (cd); Linear momentum matches with Derived quantity expressed in kg·m·s⁻¹; Thermodynamic temperature matches with Fundamental quantity measured in kelvin (K); Specific heat capacity matches with Derived quantity expressed in m²·s⁻²·K⁻¹.
Luminous intensity and thermodynamic temperature are fundamental physical quantities with SI units candela (cd\text{cd}) and kelvin (K\text{K}) respectively. Linear momentum (p=mvp = mv) and specific heat capacity (c=QmΔTc = \frac{Q}{m \Delta T}) are derived physical quantities whose fundamental SI base unit resolutions are kgms1\text{kg}\cdot\text{m}\cdot\text{s}^{-1} and m2s2K1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1} respectively.

Adım Adım Çözüm

1
Identify fundamental physical quantities and their base SI units
Luminous intensity (measured in cd\text{cd}) and thermodynamic temperature (measured in K\text{K}) are base physical quantities that cannot be expressed in terms of other quantities.
The standard SI system establishes 7 base independent quantities.
2
Resolve derived quantities into fundamental SI base units
Linear momentum (p=mvp = mv) has units kgms1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}. Specific heat capacity (c=QmΔTc = \frac{Q}{m\Delta T}) has units kgm2s2kgK=m2s2K1\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Derived physical quantities are formed by combining fundamental quantities algebraically according to physical laws.
3
Match each physical quantity to its correct description
Luminous intensity \rightarrow candela (cd\text{cd}); Linear momentum \rightarrow kgms1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}; Thermodynamic temperature \rightarrow kelvin (K\text{K}); Specific heat capacity \rightarrow m2s2K1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Each pair directly aligns the physical quantity with its base classification and unit derivation.

Anahtar Kavram

Classification of fundamental and derived physical quantities and their resolution into SI base units
Soru 7999Soru

Monochromatic radiation carrying photons of energy 4.8 eV4.8\text{ eV} illuminates a cesium surface inside a photoelectric cell. If the work function of cesium is 2.1 eV2.1\text{ eV}, determine the stopping potential, in volts, needed to reduce the photoelectric current to zero.

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Cevap: 2.7

Cevap

The stopping potential required to reduce the photoelectric current to zero is 2.7 V2.7\text{ V}.
Einstein's photoelectric equation states that incident photon energy EE equals the work function W0W_0 plus the maximum kinetic energy KmaxK_{\text{max}} of the photoelectrons (E=W0+KmaxE = W_0 + K_{\text{max}}). Rearranging gives Kmax=4.8 eV2.1 eV=2.7 eVK_{\text{max}} = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}. Since Kmax=eVsK_{\text{max}} = e V_s, an electron-volt value of kinetic energy numerically equals the stopping potential in volts, giving a stopping potential of 2.7 V2.7\text{ V}.

Adım Adım Çözüm

1
Calculate the maximum kinetic energy (KmaxK_{\text{max}}) of the emitted photoelectrons.
Kmax=EW0=4.8 eV2.1 eV=2.7 eVK_{\text{max}} = E - W_0 = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}
According to Einstein's photoelectric equation, incident photon energy is divided into overcoming the work function of the metal and providing kinetic energy to the liberated electron.
2
Determine the stopping potential (VsV_s) from the maximum kinetic energy.
Vs=Kmaxe=2.7 eVe=2.7 VV_s = \frac{K_{\text{max}}}{e} = \frac{2.7\text{ eV}}{e} = 2.7\text{ V}
The stopping potential VsV_s is the opposing potential difference needed to stop the fastest moving photoelectrons, defined by Kmax=eVsK_{\text{max}} = e V_s.

Anahtar Kavram

Photoelectric Effect and Work Function
Soru 8000Soru

A binary operation Δ\Delta defined on the set of real numbers R\mathbb{R} is given by aΔb=a+bab4a \Delta b = a + b - \frac{ab}{4}. What is the inverse of 22 under this operation?

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Cevap: 4-4

Cevap

The inverse of 22 under the binary operation is 4-4.
To find the inverse of 22, we first determine the identity element ee using aΔe=aa \Delta e = a, which yields e=0e = 0. Setting 2Δx=02 \Delta x = 0 gives 2+x2x4=02 + x - \frac{2x}{4} = 0, which simplifies to 2+x2=02 + \frac{x}{2} = 0, resulting in x=4x = -4.

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1
Determine the identity element ee of the operation.
e=0e = 0
By definition, aΔe=aa \Delta e = a. Thus, a+eae4=a    e(1a4)=0a + e - \frac{ae}{4} = a \implies e\left(1 - \frac{a}{4}\right) = 0, which gives e=0e = 0 for all real aa.
2
Set up the inverse equation for the element 22.
2+x2x4=02 + x - \frac{2x}{4} = 0
Let xx be the inverse of 22. By definition, 2Δx=e2 \Delta x = e, where e=0e = 0.
3
Simplify and solve for xx.
x=4x = -4
Simplify 2+xx2=0    2+x2=0    x2=2    x=42 + x - \frac{x}{2} = 0 \implies 2 + \frac{x}{2} = 0 \implies \frac{x}{2} = -2 \implies x = -4.

Anahtar Kavram

Inverse Element in Binary Operations
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