Tüm alıştırma soruları

13931 soru

Soru 4341Soru

An electric train starts from rest and accelerates uniformly at 2 m/s22\text{ m/s}^2 for a time interval tt. It then maintains the maximum velocity attained for a time interval of tt. Finally, it decelerates uniformly at 4 m/s24\text{ m/s}^2 until it comes to a complete stop. If the total distance covered during the entire journey is 350 m350\text{ m}, what is the value of tt?

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Cevap: 10 s10\text{ s}

Cevap

10 s10\text{ s}
The motion consists of three stages: acceleration from rest (s1=t2s_1 = t^2), uniform velocity (s2=2t2s_2 = 2t^2), and deceleration to rest (s3=0.5t2s_3 = 0.5t^2). Adding these gives a total distance of S=3.5t2S = 3.5t^2. Setting 3.5t2=350 m3.5t^2 = 350\text{ m} yields t2=100t^2 = 100, so t=10 st = 10\text{ s}.

Adım Adım Çözüm

1
Analyze Stage 1 (Acceleration phase)
Maximum velocity v=2tv = 2t, displacement s1=12(2)t2=t2s_1 = \frac{1}{2}(2)t^2 = t^2
Starting from rest (u=0u = 0) with acceleration a1=2 m/s2a_1 = 2\text{ m/s}^2 for duration tt, v=u+a1t=2tv = u + a_1 t = 2t and s1=12a1t2=t2s_1 = \frac{1}{2} a_1 t^2 = t^2.
2
Analyze Stage 2 (Constant velocity phase)
Displacement s2=(2t)(t)=2t2s_2 = (2t)(t) = 2t^2
The train moves at constant velocity v=2tv = 2t for duration tt, so distance is velocity multiplied by time.
3
Analyze Stage 3 (Deceleration phase)
Deceleration time t3=0.5tt_3 = 0.5t, displacement s3=0.5t2s_3 = 0.5t^2
Decelerating from v=2tv = 2t to 00 at a2=4 m/s2a_2 = 4\text{ m/s}^2 takes time t3=va2=2t4=0.5tt_3 = \frac{v}{a_2} = \frac{2t}{4} = 0.5t. Distance s3=12vt3=12(2t)(0.5t)=0.5t2s_3 = \frac{1}{2} v t_3 = \frac{1}{2} (2t)(0.5t) = 0.5t^2.
4
Calculate total displacement and solve for tt
Total distance S=3.5t2=350    t2=100    t=10 sS = 3.5t^2 = 350 \implies t^2 = 100 \implies t = 10\text{ s}
Summing the displacements: S=t2+2t2+0.5t2=3.5t2S = t^2 + 2t^2 + 0.5t^2 = 3.5t^2. Equating to 350 m350\text{ m} gives 3.5t2=3503.5t^2 = 350, so t2=100t^2 = 100, giving t=10 st = 10\text{ s}.

Anahtar Kavram

Multi-stage linear motion and displacement calculation
Tahmini Süre:2m 0s
Soru 4342Soru

A binary operation \ast is defined on the set of real numbers R\mathbb{R} by ab=a+bka \ast b = a + b - k, where kk is a constant. If the identity element of the operation is 44, what is the inverse of 77 under this operation?

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Cevap: 1

Cevap

The inverse of 7 under the binary operation is 1.
First, use the identity property a * e = a with e = 4 to determine that k = 4. Then, set 7 * x = 4 using the operational definition 7 + x - 4 = 4. Solving the linear equation x + 3 = 4 gives the correct inverse value of 1.

Adım Adım Çözüm

1
Determine the value of the constant k using the identity element property.
k = 4
By definition of identity element e, a * e = a. Given e = 4, substituting into the definition yields a + 4 - k = a, which simplifies to k = 4.
2
Write the full operational formula.
a * b = a + b - 4
Substitute k = 4 into the original rule a * b = a + b - k.
3
Solve for the inverse element of 7.
x = 1
Let x be the inverse of 7. By definition of inverse, 7 * x = e, so 7 + x - 4 = 4. Simplifying gives x + 3 = 4, hence x = 1.

Anahtar Kavram

Identity and inverse elements of a binary operation
Soru 4343Soru

In an agricultural survey of 100100 local farmers in a community, 5555 grow Cassava (CC), 4545 grow Yam (YY), and 4040 grow Maize (MM). It is observed that 2020 farmers grow both Cassava and Yam, 1515 grow both Yam and Maize, and 1616 grow both Cassava and Maize. If 77 farmers grow all three crops, how many farmers grow none of these three crops?

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Cevap: 44

Cevap

The number of farmers who grow none of the three crops is 4.
The number of farmers who grow at least one crop is found using n(CYM)=55+45+40201516+7=96n(C \cup Y \cup M) = 55 + 45 + 40 - 20 - 15 - 16 + 7 = 96. Subtracting this union from the total number of farmers (100100) gives 10096=4100 - 96 = 4 farmers who grow none of the three crops.

Adım Adım Çözüm

1
Apply the Principle of Inclusion-Exclusion for three sets to find the total number of farmers growing at least one crop n(CYM)n(C \cup Y \cup M).
n(CYM)=n(C)+n(Y)+n(M)n(CY)n(YM)n(CM)+n(CYM)n(C \cup Y \cup M) = n(C) + n(Y) + n(M) - n(C \cap Y) - n(Y \cap M) - n(C \cap M) + n(C \cap Y \cap M)
Elements in pairwise intersections are double-counted and elements in all three sets are triple-counted, requiring correction.
2
Substitute the given cardinalities into the formula.
n(CYM)=55+45+40201516+7=96n(C \cup Y \cup M) = 55 + 45 + 40 - 20 - 15 - 16 + 7 = 96
Calculating the total number of elements inside the union of the three sets.
3
Subtract the number of farmers in the union from the universal set size n(U)=100n(U) = 100.
n((CYM))=10096=4n((C \cup Y \cup M)') = 100 - 96 = 4
The complement of the union gives the number of elements belonging to none of the sets.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets and Universal Complement
Soru 4344Soru

A concave mirror has a radius of curvature of 40 cm40\text{ cm}. What is the focal length of the mirror?

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Cevap: 20 cm20\text{ cm}

Cevap

The focal length of the concave mirror is 20 cm20\text{ cm}.
For any spherical mirror, the focal length is half of its radius of curvature (f=r2f = \frac{r}{2}). Since the mirror is concave, its focal length is positive, yielding 40 cm2=20 cm\frac{40\text{ cm}}{2} = 20\text{ cm}.

Adım Adım Çözüm

1
Identify the given physical quantity
Radius of curvature r=40 cmr = 40\text{ cm} for a concave mirror.
The radius of curvature is the distance from the pole to the center of curvature.
2
Apply the relationship between focal length and radius of curvature
f=r2=40 cm2=20 cmf = \frac{r}{2} = \frac{40\text{ cm}}{2} = 20\text{ cm}.
For spherical mirrors, the principal focus lies halfway between the pole and the center of curvature.

Anahtar Kavram

Focal length of spherical mirrors
Soru 4345Soru

A spherical planet has a radius of 7.2×106 m7.2 \times 10^{6} \text{ m} and an acceleration due to gravity of 10 m/s210 \text{ m/s}^2 at its surface. What is the escape velocity for an object launched from the surface of this planet, in km/s\text{km/s}?

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Cevap: 12

Cevap

The escape velocity from the surface of the planet is 12 km/s.
The escape velocity vev_e from the surface of a spherical planet of radius RR with surface gravity gg is given by ve=2gRv_e = \sqrt{2gR}. Substituting g=10 m/s2g = 10 \text{ m/s}^2 and R=7.2×106 mR = 7.2 \times 10^6 \text{ m} yields ve=2×10×7.2×106=1.44×108=12000 m/sv_e = \sqrt{2 \times 10 \times 7.2 \times 10^6} = \sqrt{1.44 \times 10^8} = 12000 \text{ m/s}, which equals 12 km/s12 \text{ km/s}.

Adım Adım Çözüm

1
Identify the relationship between surface gravity, planetary radius, and escape velocity
The escape velocity formula is ve=2gRv_e = \sqrt{2gR}.
Escape velocity is the minimum initial speed required for an object to overcome the gravitational pull of a celestial body.
2
Substitute the given numerical values into the formula
ve=2×10 m/s2×7.2×106 m=144×106 m/sv_e = \sqrt{2 \times 10 \text{ m/s}^2 \times 7.2 \times 10^6 \text{ m}} = \sqrt{144 \times 10^6} \text{ m/s}.
Plugging in the given values allows direct calculation of the velocity in standard SI units.
3
Simplify the square root and convert units to km/s
ve=12000 m/s=12 km/sv_e = 12000 \text{ m/s} = 12 \text{ km/s}.
Taking the square root of 144×106144 \times 10^6 gives 12000 m/s12000 \text{ m/s}, which corresponds to 12 km/s12 \text{ km/s}.

Anahtar Kavram

Escape Velocity from a Planet's Surface
Soru 4346Soru

Match each experimental temperature measurement requirement on the left with the most appropriate thermometric instrument on the right based on its thermometric property and operational characteristics.

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Öğeler

Standard calibration reference requiring high accuracy over a wide range using pressure variations at constant volume
High-precision steady-state measurement using electrical resistance variation where slight thermal response lag is permissible
Measurement of rapidly changing temperatures at a localized point using thermal electromotive force (e.m.f.)
Non-contact measurement of extremely high temperatures of glowing bodies using radiant energy intensity

Eşleşmeler

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Cevap

The correct pairings match each measurement requirement to its corresponding thermometric instrument based on its fundamental thermometric property: standard reference calibration pairs with the constant-volume gas thermometer; high-precision steady measurement pairs with the platinum resistance thermometer; rapid localized temperature change measurement pairs with the thermocouple; and non-contact high-temperature measurement pairs with the optical pyrometer.
Each instrument is correctly matched according to the specific physical property that changes measurably with temperature (PP, RR, e.m.f., and radiation intensity) and its operational suitability.

Adım Adım Çözüm

1
Analyze requirement 1: standard reference calibration using gas pressure at constant volume.
Identified thermometric property as pressure PP at constant volume VV, which defines the constant-volume gas thermometer.
Gas thermometers closely approximate the absolute thermodynamic scale and serve as calibration standards.
2
Analyze requirement 2: high-precision steady measurement using resistance variation with thermal lag.
Identified thermometric property as electrical resistance RR, which corresponds to the platinum resistance thermometer.
Platinum wire resistance changes predictably with temperature, providing high accuracy for stable temperatures.
3
Analyze requirement 3: rapid localized temperature measurement via thermal e.m.f.
Identified thermometric property as thermoelectric voltage (e.m.f.), which corresponds to the thermocouple.
The small thermal mass of thermocouple junctions allows low response times for fast transient measurements.
4
Analyze requirement 4: non-contact measurement of glowing bodies using radiation.
Identified physical principle as thermal radiation intensity, corresponding to the optical pyrometer.
Pyrometers detect infrared/visible radiation, avoiding structural melting associated with direct contact at extreme temperatures.

Anahtar Kavram

Thermometric Properties and Operational Limits of Thermometers
Tahmini Süre:2m 0s
Soru 4347Soru

An acoustic pulse generator located at a fixed point between two parallel rigid reflective barriers emits a sound wave with a frequency of 680 Hz680\text{ Hz}. The first echo from the closer barrier is detected after 0.8 s0.8\text{ s}, and the first echo from the farther barrier is detected after 1.4 s1.4\text{ s}. Assuming the speed of sound in the medium is 340 m s1340\text{ m s}^{-1}, what is the total distance between the two barriers, and how many complete wavelengths of this sound wave fit within this total distance?

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Cevap: 374 m374\text{ m} and 748748 wavelengths

Cevap

The total distance between the barriers is 374 m374\text{ m} and 748748 complete wavelengths fit within this distance.
The distance to the first barrier is d1=340×0.82=136 md_1 = \frac{340 \times 0.8}{2} = 136\text{ m}, and to the second barrier is d2=340×1.42=238 md_2 = \frac{340 \times 1.4}{2} = 238\text{ m}. Adding both distances gives a total separation of 374 m374\text{ m}. With a wavelength λ=vf=340680=0.5 m\lambda = \frac{v}{f} = \frac{340}{680} = 0.5\text{ m}, the number of full wavelengths fitting in 374 m374\text{ m} is 3740.5=748\frac{374}{0.5} = 748.

Adım Adım Çözüm

1
Calculate the one-way distance from the generator to each reflective barrier using the echo relationship d=vt2d = \frac{v \cdot t}{2}.
For barrier 1: d1=340 m s1×0.8 s2=136 md_1 = \frac{340 \text{ m s}^{-1} \times 0.8 \text{ s}}{2} = 136 \text{ m}. For barrier 2: d2=340 m s1×1.4 s2=238 md_2 = \frac{340 \text{ m s}^{-1} \times 1.4 \text{ s}}{2} = 238 \text{ m}.
An echo represents a two-way journey (to the surface and back), so the time taken to travel the one-way distance is half of the echo reception time.
2
Determine the total distance between the two parallel barriers.
D=d1+d2=136 m+238 m=374 mD = d_1 + d_2 = 136 \text{ m} + 238 \text{ m} = 374 \text{ m}.
Since the generator is situated between the two barriers, the total separation distance equals the sum of the individual distances to each barrier.
3
Calculate the wavelength λ\lambda of the sound wave using the wave equation v=fλv = f \lambda.
\lambda = \frac{v}{f} = \frac{340 \text{ m s}^{-1}}{680 \text{ Hz}} = 0.5 \text{ m}$.
Wavelength is the ratio of wave speed to frequency.
4
Compute the number of complete wavelengths NN contained within the total separation distance.
N = \frac{D}{\lambda} = \frac{374 \text{ m}}{0.5 \text{ m}} = 748.
Dividing total distance by single wavelength yields the number of full wave cycles occupying that span.

Anahtar Kavram

Echo distance calculations and wave speed-frequency-wavelength relationships
Tahmini Süre:2m 0s
Soru 4348Soru

A flexible balloon contains a sample of gas occupying a volume of 3.0 m33.0\text{ m}^3 at a temperature of 27C27^\circ\text{C} under constant pressure. If the gas is heated to 127C127^\circ\text{C} while keeping the pressure constant, what is the new volume of the gas in m3\text{m}^3?

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Cevap: 4

Cevap

The new volume of the gas is 4.0 m34.0\text{ m}^3.
According to Charles's Law (V1T1=V2T2 \frac{V_1}{T_1} = \frac{V_2}{T_2}), at constant pressure, volume is directly proportional to absolute temperature in Kelvin. Converting 27C27^\circ\text{C} to 300 K300\text{ K} and 127C127^\circ\text{C} to 400 K400\text{ K} yields a final volume V2=3.0×400300=4.0 m3V_2 = 3.0 \times \frac{400}{300} = 4.0\text{ m}^3.

Adım Adım Çözüm

1
Convert the given initial and final temperatures from Celsius to Kelvin
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=127+273=400 KT_2 = 127 + 273 = 400\text{ K}
Gas law calculations require absolute temperature in Kelvin.
2
Set up Charles's Law equation for constant pressure processes
V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}
At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.
3
Substitute the values into the equation and solve for the final volume V2V_2
V2=3.0×400 K300 K=4.0 m3V_2 = 3.0 \times \frac{400\text{ K}}{300\text{ K}} = 4.0\text{ m}^3
Direct algebraic evaluation yields the final gas volume.

Anahtar Kavram

Charles's Law
Soru 4349Soru

A straight conductor of length 0.5 m0.5\text{ m} carries a current of 4.0 A4.0\text{ A} in a uniform magnetic field of flux density 0.20 T0.20\text{ T}. If the conductor is placed at an angle of 3030^\circ to the magnetic field lines, what is the magnitude of the magnetic force exerted on the conductor?

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Cevap: 0.20 N0.20\text{ N}

Cevap

0.20 N0.20\text{ N}
The magnetic force on a straight wire carrying current in a uniform magnetic field is given by F=BILsinθF = BIL \sin\theta. Substituting the given values B=0.20 TB = 0.20\text{ T}, I=4.0 AI = 4.0\text{ A}, L=0.5 mL = 0.5\text{ m}, and θ=30\theta = 30^\circ yields F=0.20×4.0×0.5×0.5=0.20 NF = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}.

Adım Adım Çözüm

1
Identify the given physical quantities
L=0.5 mL = 0.5\text{ m}, I=4.0 AI = 4.0\text{ A}, B=0.20 TB = 0.20\text{ T}, and θ=30\theta = 30^\circ.
Extracting values from the problem statement ensures correct substitution.
2
State the magnetic force equation for a straight current-carrying conductor in a magnetic field
F=BILsinθF = BIL \sin\theta
Only the component of the magnetic field perpendicular to the current contributes to the magnetic force.
3
Substitute the values and calculate the force magnitude
F=(0.20 T)×(4.0 A)×(0.5 m)×sin(30)=0.20×4.0×0.5×0.5=0.20 NF = (0.20\text{ T}) \times (4.0\text{ A}) \times (0.5\text{ m}) \times \sin(30^\circ) = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}.
Evaluating sin30=0.5\sin 30^\circ = 0.5 gives the required force value.

Anahtar Kavram

Magnetic Force on a Current-Carrying Conductor
Tahmini Süre:1m 15s
Soru 4350Soru

1010 distinct points are marked on the circumference of a circle. How many different triangles can be formed by connecting any 33 of these points as vertices?

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Cevap: 120120

Cevap

The number of distinct triangles that can be formed is 120120.
To form a triangle, any 33 points must be chosen from the 1010 available points. Because no three points on a circle are collinear, every choice of 33 points forms a unique triangle. Since the order of choosing vertices does not alter the triangle, we use combinations: (103)=10×9×83×2×1=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.

Adım Adım Çözüm

1
Identify the total number of items nn and the subset size rr.
n=10n = 10 points on the circle, and r=3r = 3 points required to form a triangle.
Any set of 33 non-collinear points uniquely determines a triangle. Since all points lie on a circle, no three points are collinear.
2
Determine whether order matters.
Order does not matter because choosing points A,B,CA, B, C produces the same triangle as choosing B,C,AB, C, A.
Selection of vertices for a geometric shape is a combination problem, not a permutation problem.
3
Apply the combination formula (nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n - r)!}.
(103)=10×9×83×2×1=7206=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120
Evaluating the combination gives the total number of distinct triangles.

Anahtar Kavram

Combinations for Geometric Formations
Tahmini Süre:1m 30s
Soru 4351Soru

Two long parallel straight conductors separated by a distance of 0.20 m0.20\text{ m} in vacuum carry steady currents of 5.0 A5.0\text{ A} and 8.0 A8.0\text{ A} in opposite directions. Given that the permeability of free space μ0=4π×107 Tm/A\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}, what is the magnitude of the magnetic force per unit length acting between the conductors and the nature of this force?

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Cevap: 4.0×105 N/m4.0 \times 10^{-5}\text{ N/m}, repulsive

Cevap

4.0×105 N/m4.0 \times 10^{-5}\text{ N/m}, repulsive
Applying the formula for force per unit length gives FL=μ0I1I22πd=(4π×107)(5.0)(8.0)2π(0.20)=4.0×105 N/m\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(5.0)(8.0)}{2\pi (0.20)} = 4.0 \times 10^{-5}\text{ N/m}. Since the currents flow in opposite directions, the magnetic interaction is repulsive.

Adım Adım Çözüm

1
Identify given variables and the formula for force per unit length between parallel conductors.
Given: I1=5.0 AI_1 = 5.0\text{ A}, I2=8.0 AI_2 = 8.0\text{ A}, d=0.20 md = 0.20\text{ m}, μ0=4π×107 Tm/A\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}. Formula: FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}.
The magnetic force per unit length between two parallel wires depends on the current magnitudes and their distance of separation.
2
Substitute values into the equation and compute the magnitude.
FL=(4π×107)(5.0)(8.0)2π(0.20)=2×107×400.20=4.0×105 N/m\frac{F}{L} = \frac{(4\pi \times 10^{-7})(5.0)(8.0)}{2\pi (0.20)} = \frac{2 \times 10^{-7} \times 40}{0.20} = 4.0 \times 10^{-5}\text{ N/m}.
Canceling 4π4\pi with 2π2\pi yields a multiplier of 2×1072 \times 10^{-7}, simplifying numerical computation.
3
Determine the direction of force based on current flow direction.
The force is repulsive.
Parallel conductors carrying currents in opposite directions generate magnetic field configurations that force the wires apart.

Anahtar Kavram

Magnetic force per unit length between parallel current-carrying conductors
Soru 4352Soru

A toy truck of mass 2.5 kg2.5\text{ kg} moving at a velocity of 6.0 m s16.0\text{ m s}^{-1} collides head-on with a toy car of mass 1.5 kg1.5\text{ kg} moving in the opposite direction at 2.0 m s12.0\text{ m s}^{-1}. If the two vehicles stick together upon impact, what is their combined velocity immediately after the collision?

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Cevap: 3.0 m s13.0\text{ m s}^{-1}

Cevap

3.0 m s13.0\text{ m s}^{-1} in the direction of the initial motion of the truck
According to the principle of conservation of linear momentum, total initial momentum equals total final momentum. Taking the direction of the toy truck as positive gives an initial momentum of (2.5 kg×6.0 m s1)+(1.5 kg×2.0 m s1)=15.03.0=12.0 kg m s1(2.5\text{ kg} \times 6.0\text{ m s}^{-1}) + (1.5\text{ kg} \times -2.0\text{ m s}^{-1}) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}. Dividing this net momentum by the combined mass (2.5 kg+1.5 kg=4.0 kg)(2.5\text{ kg} + 1.5\text{ kg} = 4.0\text{ kg}) yields a common final velocity of 3.0 m s13.0\text{ m s}^{-1}.

Adım Adım Çözüm

1
Assign direction signs to the velocity vectors
Let the initial direction of the truck be positive (u1=+6.0 m s1u_1 = +6.0\text{ m s}^{-1}). The car moves in the opposite direction, so its velocity is negative (u2=2.0 m s1u_2 = -2.0\text{ m s}^{-1}).
Linear momentum is a vector quantity, so direction must be accounted for in one-dimensional motion.
2
Calculate the total initial momentum of the system
pinitial=m1u1+m2u2=(2.5×6.0)+(1.5×2.0)=15.03.0=12.0 kg m s1p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.5 \times 6.0) + (1.5 \times -2.0) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}
The total momentum before collision is the vector sum of individual momenta.
3
Apply the law of conservation of linear momentum to solve for common final velocity vv
pfinal=(m1+m2)v=(2.5+1.5)v=4.0vp_{\text{final}} = (m_1 + m_2) v = (2.5 + 1.5) v = 4.0 v. Setting pfinal=pinitial4.0v=12.0v=3.0 m s1p_{\text{final}} = p_{\text{initial}} \Rightarrow 4.0 v = 12.0 \Rightarrow v = 3.0\text{ m s}^{-1}.
Since the vehicles stick together, they move as a single combined mass.

Anahtar Kavram

Conservation of Linear Momentum in Inelastic Collisions
Tahmini Süre:50s
Soru 4353Soru

A meteorological balloon filled with 0.50 kg0.50\text{ kg} of helium gas is launched at sea level, where the atmospheric pressure is 1.01×105 Pa1.01 \times 10^5\text{ Pa} and the ambient temperature is 27C27^\circ\text{C}. The balloon ascends to a high altitude where the external ambient pressure decreases to 4.00×104 Pa4.00 \times 10^4\text{ Pa} and the ambient temperature drops to 23C-23^\circ\text{C}. As the balloon expands, its elastic membrane exerts an additional pressure, causing the internal gas pressure to be 20%20\% higher than the surrounding ambient pressure. Assuming helium behaves as an ideal gas with a molar mass of 4.0 g/mol4.0\text{ g/mol} and the molar gas constant R=8.31 J mol1K1R = 8.31\text{ J mol}^{-1}\text{K}^{-1}, calculate the final volume of helium gas inside the balloon at this altitude in m3\text{m}^3.

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Cevap: 5.41

Cevap

The final volume of helium gas inside the balloon at altitude is 5.41 m35.41\text{ m}^3.
The ideal gas equation PV=nRTPV = nRT relates state variables. By determining n=125 molesn = 125\text{ moles} from mass and molar mass, absolute temperature T2=250 KT_2 = 250\text{ K}, and total internal pressure P2=4.80×104 PaP_2 = 4.80 \times 10^4\text{ Pa}, the final volume V2V_2 evaluates to 5.41 m35.41\text{ m}^3.

Adım Adım Çözüm

1
Convert the mass of helium into moles
n=125 molesn = 125\text{ moles}
Mass m=0.50 kg=500 gm = 0.50\text{ kg} = 500\text{ g} divided by molar mass M=4.0 g/molM = 4.0\text{ g/mol} gives n=5004.0=125 molesn = \frac{500}{4.0} = 125\text{ moles}.
2
Convert final temperature to Kelvin
T2=250 KT_2 = 250\text{ K}
Gas laws strictly require thermodynamic temperature: T2=23+273=250 KT_2 = -23 + 273 = 250\text{ K}.
3
Calculate final internal pressure of the gas
P2=4.80×104 PaP_2 = 4.80 \times 10^4\text{ Pa}
The gas pressure inside the balloon is 20%20\% higher than external ambient pressure: P2=1.20×(4.00×104)=4.80×104 PaP_2 = 1.20 \times (4.00 \times 10^4) = 4.80 \times 10^4\text{ Pa}.
4
Solve for the final volume using the ideal gas equation
V2=5.41 m3V_2 = 5.41\text{ m}^3
Rearranging P2V2=nRT2P_2 V_2 = n R T_2 yields V2=nRT2P2=125×8.31×2504.80×104=5.41015... m35.41 m3V_2 = \frac{n R T_2}{P_2} = \frac{125 \times 8.31 \times 250}{4.80 \times 10^4} = 5.41015...\text{ m}^3 \approx 5.41\text{ m}^3.

Anahtar Kavram

Ideal Gas Equation (PV=nRTPV = nRT)
Tahmini Süre:3m 0s
Soru 4354Soru

A water tank has a real depth of 16.0 cm16.0\text{ cm}. If the refractive index of water relative to air is 43\frac{4}{3}, what is the apparent depth of the tank in centimetres when viewed normally from above?

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Cevap: 12

Cevap

The apparent depth of the water tank is 12.0 cm12.0\text{ cm}.
When light passes from water to air, refraction causes the apparent depth to be smaller than the real depth by a factor equal to the refractive index nn. Substituting a real depth of 16.0 cm16.0\text{ cm} and n=43n = \frac{4}{3} into Apparent Depth=Real Depthn\text{Apparent Depth} = \frac{\text{Real Depth}}{n} yields 12.0 cm12.0\text{ cm}.

Adım Adım Çözüm

1
Identify the relationship between refractive index, real depth, and apparent depth.
n=Real DepthApparent Depthn = \frac{\text{Real Depth}}{\text{Apparent Depth}}
Refraction at a plane boundary causes an object immersed in a denser medium to appear closer to the surface when viewed from a rarer medium.
2
Rearrange the equation to isolate Apparent Depth.
Apparent Depth=Real Depthn\text{Apparent Depth} = \frac{\text{Real Depth}}{n}
The unknown quantity requested by the question is the apparent depth.
3
Substitute the known numerical values into the equation.
Apparent Depth=16.043=16.0×34=12.0 cm\text{Apparent Depth} = \frac{16.0}{\frac{4}{3}} = 16.0 \times \frac{3}{4} = 12.0\text{ cm}
Dividing 16.016.0 by 43\frac{4}{3} gives 12.012.0.

Anahtar Kavram

Real and Apparent Depth in Light Refraction
Soru 4355Soru

A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end records two sound signals—one traveling through the aluminum pipeline and the other through the surrounding air—separated by a time interval of 2.8 s2.8\text{ s}. If the speed of sound in air is 340 m s1340\text{ m s}^{-1} and the speed of sound in aluminum is 5100 m s15100\text{ m s}^{-1}, what is the length of the pipeline in meters?

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Cevap: 1020

Cevap

The length of the pipeline is 1020 m1020\text{ m}.
Sound travels significantly faster through solids like aluminum (5100 m s15100\text{ m s}^{-1}) than through gases like air (340 m s1340\text{ m s}^{-1}). The time taken for sound to travel a distance LL through air is tair=L340t_{\text{air}} = \frac{L}{340}, while through aluminum it is tmetal=L5100t_{\text{metal}} = \frac{L}{5100}. Setting their difference equal to 2.8 s2.8\text{ s} gives L340L5100=2.8\frac{L}{340} - \frac{L}{5100} = 2.8, which solves to L=1020 mL = 1020\text{ m}.

Adım Adım Çözüm

1
Formulate transit time expressions for both media
tair=L340t_{\text{air}} = \frac{L}{340} and tmetal=L5100t_{\text{metal}} = \frac{L}{5100}
Time taken by a wave to travel distance LL at constant speed vv is t=Lvt = \frac{L}{v}.
2
Set up the time difference equation
tairtmetal=2.8 st_{\text{air}} - t_{\text{metal}} = 2.8\text{ s}
The sound wave travels faster through aluminum than air, so the air pulse arrives later by 2.8 s2.8\text{ s}.
3
Solve the algebraic equation for distance LL
L=1020 mL = 1020\text{ m}
Combining terms yields 14L5100=2.8\frac{14L}{5100} = 2.8, which simplifies to L=2.8×510014=1020 mL = \frac{2.8 \times 5100}{14} = 1020\text{ m}.

Anahtar Kavram

Propagation speed of sound waves in different physical media
Soru 4356Soru

According to the kinetic theory of matter, what causes the pressure exerted by a gas on the walls of its container?

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Cevap: The continuous elastic collisions of gas molecules with the walls of the container

Cevap

The pressure exerted by a gas is due to the continuous elastic collisions of gas molecules with the container walls.
According to the kinetic theory of gases, gas particles are in rapid, random motion. When these particles collide elastically with the walls of the container, they undergo a change in momentum. The average rate of momentum change per unit area exerted by countless particle impacts manifests as macroscopic gas pressure.

Adım Adım Çözüm

1
Identify the basic postulate of the kinetic theory of gases regarding particle motion.
Gas molecules are in rapid, constant, and random motion.
Kinetic theory models gas behavior based on particle movement.
2
Relate particle motion to force and pressure on the container boundary.
When particles hit the container wall, they undergo a change in momentum, imparting a force on the wall.
Force is defined as the rate of change of momentum (F=ΔpΔtF = \frac{\Delta p}{\Delta t}).
3
Define pressure in terms of force per unit surface area.
The total force exerted per unit surface area by continuous collisions results in gas pressure (P=FAP = \frac{F}{A}).
Pressure is force distributed over a given surface area.

Anahtar Kavram

Kinetic Theory of Matter and Pressure of Gases
Soru 4357Soru

Match each electromagnetic device or component listed on the left with its corresponding function or operating principle on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Moving coil galvanometer
Electric motor
Split-ring commutator
Soft iron core

Eşleşmeler

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Cevap

Moving coil galvanometer matches with detecting/measuring small currents via torque; Electric motor matches with converting electrical energy to mechanical energy; Split-ring commutator matches with reversing current every half-cycle to maintain continuous rotation; Soft iron core matches with concentrating magnetic flux to create a radial magnetic field.
Each electromagnetic device or component relies on magnetic forces: moving coil galvanometers convert current to proportional coil torque; electric motors convert electrical energy into continuous mechanical rotation; split-ring commutators reverse current every half-turn to keep torque unidirectionally directed; soft iron cores concentrate magnetic flux to establish strong radial fields.

Adım Adım Çözüm

1
Identify the primary operational application of the moving coil galvanometer.
It detects small currents using the torque T=NIABsinθT = NIAB \sin\theta exerted on a current-carrying coil in a magnetic field.
This establishes the link between galvanic deflection and electrical current measurement.
2
Identify the energy transformation principle of an electric motor.
The motor converts input electrical power into mechanical torque via magnetic force F=BILsinθF = BIL\sin\theta.
This connects the motor to its fundamental mechanical output function.
3
Analyze the mechanical role of a split-ring commutator in DC devices.
It alternates current flow directions in the loop every 180180^\circ.
Without reversals, the coil would oscillate around equilibrium instead of continuously rotating.
4
Determine the ferromagnetic enhancement provided by a soft iron core.
It increases field strength and maintains a radial field orientation.
High magnetic permeability concentrates magnetic field lines effectively.

Anahtar Kavram

Operating principles and structural functions of electromagnetic devices based on magnetic torque and forces
Soru 4358Soru

An aircraft flies due East at a velocity of 120 m/s120\text{ m/s} while a crosswind blows due North at 50 m/s50\text{ m/s}. What is the magnitude of the resultant velocity of the aircraft?

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Cevap: 130 m/s130\text{ m/s}

Cevap

The magnitude of the resultant velocity of the aircraft is 130 m/s130\text{ m/s}.
Because the aircraft's velocity and the wind's velocity are perpendicular to each other, their vector sum is found using the Pythagorean theorem R=v12+v22R = \sqrt{v_1^2 + v_2^2}. Substituting 120 m/s120\text{ m/s} and 50 m/s50\text{ m/s} gives R=1202+502=16900=130 m/sR = \sqrt{120^2 + 50^2} = \sqrt{16900} = 130\text{ m/s}.

Adım Adım Çözüm

1
Identify the given velocity vectors and their relative direction.
Velocity due East ve=120 m/s\vec{v}_e = 120\text{ m/s} and velocity due North vn=50 m/s\vec{v}_n = 50\text{ m/s} are perpendicular to each other (9090^\circ angle).
East and North directions are orthogonal to each other.
2
Apply the Pythagorean theorem to calculate the magnitude of the resultant vector RR.
R=ve2+vn2=1202+502=14400+2500=16900=130 m/sR = \sqrt{v_e^2 + v_n^2} = \sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\text{ m/s}.
The resultant of two perpendicular vectors forms the hypotenuse of a right-angled triangle.

Anahtar Kavram

Vector Addition of Perpendicular Vectors
Tahmini Süre:1m 0s
Soru 4359Soru

If (x,y)(x, y) satisfies the simultaneous equations xy=1x - y = 1 and x2+y2=25x^2 + y^2 = 25, what is the value of the product xyxy?

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Cevap: 12

Cevap

12
Expanding (xy)2(x - y)^2 yields x22xy+y2x^2 - 2xy + y^2. Substituting the given values xy=1x - y = 1 and x2+y2=25x^2 + y^2 = 25 into this identity gives 1=252xy1 = 25 - 2xy. Rearranging yields 2xy=242xy = 24, which solves to xy=12xy = 12. Alternatively, solving by substitution gives solution pairs (4,3)(4, 3) and (3,4)(-3, -4), both yielding a product of 1212.

Adım Adım Çözüm

1
Apply the algebraic expansion identity
(xy)2=x2+y22xy(x - y)^2 = x^2 + y^2 - 2xy
Connects the difference of terms, the sum of their squares, and their product.
2
Substitute the values given in the system of equations
12=252xy1^2 = 25 - 2xy
Replaces xyx - y with 1 and x2+y2x^2 + y^2 with 25.
3
Isolate and calculate the product xyxy
2xy=24    xy=122xy = 24 \implies xy = 12
Simplifies 1=252xy1 = 25 - 2xy to find the exact numerical value of xyxy.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
Soru 4360Soru

A projectile is launched from ground level over flat terrain with a constant horizontal velocity component of 10 m/s10\text{ m/s}. At a height of 40 m40\text{ m} above the ground, the magnitude of its vertical velocity component is equal to its horizontal velocity component. Taking g=10 m/s2g = 10\text{ m/s}^2, calculate the maximum height reached by the projectile in meters.

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Cevap: 45

Cevap

The maximum height reached by the projectile is 45 m45\text{ m}.
At the height of 40 m40\text{ m}, the vertical velocity is equal to the horizontal velocity of 10 m/s10\text{ m/s}. Applying the vertical kinematic relation vy2=uy22ghv_y^2 = u_y^2 - 2gh yields 102=uy22(10)(40)10^2 = u_y^2 - 2(10)(40), which gives uy2=900 m2/s2u_y^2 = 900\text{ m}^2/\text{s}^2. The maximum height attained above ground level occurs when vy=0v_y = 0, calculated as Hmax=uy22g=90020=45 mH_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}.

Adım Adım Çözüm

1
Identify the vertical component of velocity at the given height
vy=10 m/sv_y = 10\text{ m/s} at h=40 mh = 40\text{ m}
The problem states that at h=40 mh = 40\text{ m}, the vertical velocity component equals the constant horizontal component ux=10 m/su_x = 10\text{ m/s}.
2
Determine the initial vertical launch velocity component uyu_y
uy2=900 m2/s2    uy=30 m/su_y^2 = 900\text{ m}^2/\text{s}^2 \implies u_y = 30\text{ m/s}
Applying the vertical motion equation vy2=uy22ghv_y^2 = u_y^2 - 2gh gives 102=uy22(10)(40)    uy2=100+800=90010^2 = u_y^2 - 2(10)(40) \implies u_y^2 = 100 + 800 = 900.
3
Calculate the maximum height HmaxH_{\text{max}}
Hmax=45 mH_{\text{max}} = 45\text{ m}
At maximum height, the vertical velocity becomes zero. Using Hmax=uy22g=90020=45 mH_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}.

Anahtar Kavram

Vertical Kinematics and Maximum Height of a Projectile
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