Tüm alıştırma soruları

Soru 1501Soru →

A sealed rigid glass bulb contains a fixed mass of helium gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Assuming the volume of the bulb remains constant, calculate the final pressure of the gas in $\text{Pa}$ when it is heated to $127^\circ\text{C}$ .

Cevabı ve açıklamayı göster

Cevap: 160000

Cevap

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

(

160\,000\text{ Pa}

).

The correct result of

160\,000\text{ Pa}

(

1.60 \times 10^5\text{ Pa}

) is obtained by applying the Pressure Law

\frac{P_1}{T_1} = \frac{P_2}{T_2}

using absolute temperatures (

T_1 = 300\text{ K}

,

T_2 = 400\text{ K}

).

Adım Adım Çözüm

1

Convert the initial and final temperatures from degrees Celsius to the absolute temperature scale (Kelvin).

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

.

All gas law equations require absolute temperatures in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) for a fixed mass of gas at constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

The pressure of a gas is directly proportional to its absolute temperature when volume remains constant.

3

Substitute the known values to find

P_2

.

P_2 = 1.20 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa} = 160\,000\text{ Pa}

.

Multiplying

1.20 \times 10^5

by the temperature ratio

\frac{4}{3}

gives

1.60 \times 10^5\text{ Pa}

.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law) and absolute temperature conversion

Soru 1502Soru →

A rigid container $A$ of volume $0.060\text{ m}^3$ contains an ideal gas at an absolute pressure of $4.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Container $A$ is connected via a narrow tube of negligible volume with a valve to an evacuated rigid container $B$ of volume $0.040\text{ m}^3$ . The valve is opened, allowing gas to flow between the containers until equilibrium is established. If container $A$ is maintained at $27^\circ\text{C}$ while container $B$ is maintained at $127^\circ\text{C}$ , what is the final equilibrium pressure of the gas in kilopascals (kPa)?

Cevabı ve açıklamayı göster

Cevap: 300

Cevap

300 kPa

Converting both temperatures to Kelvin (

300\text{ K}

and

400\text{ K}

) and applying mole conservation (

n_{\text{total}} = n_A + n_B

) yields a final uniform pressure of

3.00 \times 10^5\text{ Pa}

, which equals

300\text{ kPa}

.

Adım Adım Çözüm

1

Convert temperatures from degrees Celsius to kelvins.

T_A = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_B = 127^\circ\text{C} + 273 = 400\text{ K}

.

Gas laws and calculations using the ideal gas equation require absolute temperatures in Kelvin.

2

Calculate the initial number of moles of gas present in the system.

n_{\text{total}} = \frac{p_A V_A}{R T_A} = \frac{(4.50 \times 10^5\text{ Pa})(0.060\text{ m}^3)}{R(300\text{ K})} = \frac{90}{R}

.

Container B is initially evacuated, meaning all gas molecules originate from container A.

3

Express the total number of moles at final equilibrium in terms of final pressure

P

.

n_{\text{final}} = n_A + n_B = \frac{P V_A}{R T_A} + \frac{P V_B}{R T_B} = \frac{P}{R} \left(\frac{0.060}{300} + \frac{0.040}{400}\right) = \frac{P}{R} (2.0 \times 10^{-4} + 1.0 \times 10^{-4}) = \frac{3.0 \times 10^{-4} P}{R}

.

At equilibrium, the pressure

P

becomes uniform throughout both interconnected containers.

4

Equate the initial and final total moles to solve for the equilibrium pressure

P

.

\frac{90}{R} = \frac{3.0 \times 10^{-4} P}{R} \implies P = \frac{90}{3.0 \times 10^{-4}} = 3.00 \times 10^5\text{ Pa} = 300\text{ kPa}

.

The total mass and number of moles of gas are conserved within the sealed system.

Anahtar Kavram

Conservation of total moles in interconnected gas containers governed by the ideal gas equation

PV = nRT

Tahmini Süre:2m 30s

Soru 1503Soru →

An air bubble with an initial volume of $4.0\text{ cm}^3$ is released by a scuba diver at a depth where the total pressure is $2.50 \times 10^5\text{ Pa}$ and the water temperature is $7^\circ\text{C}$ . Calculate the volume of the bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the water temperature is $77^\circ\text{C}$ .

Cevabı ve açıklamayı göster

Cevap: 12.5

Cevap

12.5 cm³

Converting temperatures to the Kelvin scale (

T_1 = 280\text{ K}

and

T_2 = 350\text{ K}

) and applying the combined gas law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

gives a final volume of

12.5\text{ cm}^3

.

Adım Adım Çözüm

1

Convert the initial and final temperatures from degrees Celsius to absolute temperatures in Kelvin.

T_1 = 7^\circ\text{C} + 273 = 280\text{ K}

and

T_2 = 77^\circ\text{C} + 273 = 350\text{ K}

.

Gas law calculations require thermodynamic temperature measured on the Kelvin scale.

2

Set up the combined gas law relationship for a fixed mass of gas.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The mass of air inside the bubble remains constant while pressure, volume, and temperature change simultaneously.

3

Substitute the known values into the equation and solve for the final volume

V_2

.

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{(2.50 \times 10^5\text{ Pa}) \times (4.0\text{ cm}^3) \times (350\text{ K})}{(1.00 \times 10^5\text{ Pa}) \times (280\text{ K})} = 12.5\text{ cm}^3

.

Algebraic substitution yields the correct final volume of the expanded bubble.

Anahtar Kavram

Combined Gas Law

Soru 1504Soru →

A transverse progressive wave traveling along a taut string is represented by the equation $y(x, t) = 0.04 \sin(200t - 8x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the maximum transverse speed of a particle on the string in meters per second?

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Cevap: 8

Cevap

The maximum transverse speed of a particle on the string is

8.0\text{ m/s}

.

The maximum speed of a particle executing simple harmonic motion as part of a progressive wave is given by

v_{p,\text{max}} = A\omega

. From the wave equation

y(x, t) = 0.04 \sin(200t - 8x)

, the amplitude is

A = 0.04\text{ m}

and the angular frequency is

\omega = 200\text{ rad/s}

. Multiplying these values yields

v_{p,\text{max}} = 0.04 \times 200 = 8.0\text{ m/s}

.

Adım Adım Çözüm

1

Compare the given wave displacement equation with the standard progressive wave form.

Standard form:

y(x, t) = A \sin(\omega t - kx)

. Matching coefficients gives

A = 0.04\text{ m}

,

\omega = 200\text{ rad/s}

, and

k = 8\text{ rad/m}

.

Extracting amplitude and angular frequency is essential for evaluating particle motion.

2

Differentiate displacement with respect to time to find the expression for particle velocity.

v_p(x, t) = \frac{\partial y}{\partial t} = A \omega \cos(\omega t - kx)

.

Particle velocity represents the time rate of change of transverse displacement.

3

Determine maximum particle speed by setting the magnitude of the cosine factor to 1.

v_{p,\text{max}} = A \omega = 0.04 \text{ m} \times 200 \text{ rad/s} = 8.0 \text{ m/s}

.

The maximum absolute value of the cosine function is 1.

Anahtar Kavram

Maximum particle velocity in a progressive wave (

v_{p,\text{max}} = A\omega

)

Tahmini Süre:1m 30s

Soru 1505Soru →

A container made of a metal with a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ has a volume of $1000\text{ cm}^3$ at $20^\circ\text{C}$ . It is completely filled with a liquid at this temperature. When the system is heated to $120^\circ\text{C}$ , $40\text{ cm}^3$ of the liquid overflows. If the real cubic expansivity of the liquid is expressed as $X \times 10^{-4}\text{ K}^{-1}$ , what is the value of $X$ ?

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Cevap: 4.6

Cevap

The value of

X

is 4.6.

The real cubic expansivity of a liquid is given by

\gamma_r = \gamma_a + \gamma_v

. Calculating apparent expansivity gives

\gamma_a = \frac{40}{1000 \times 100} = 4.0 \times 10^{-4}\text{ K}^{-1}

. The vessel's volume expansivity is

\gamma_v = 3 \times 2.0 \times 10^{-5} = 0.6 \times 10^{-4}\text{ K}^{-1}

. Adding them together yields

\gamma_r = 4.6 \times 10^{-4}\text{ K}^{-1}

, so

X = 4.6

.

Adım Adım Çözüm

1

Calculate the temperature change

\Delta T

\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}

The thermal expansion is driven by the change in temperature.

2

Calculate the apparent cubic expansivity

\gamma_a

\gamma_a = \frac{40\text{ cm}^3}{1000\text{ cm}^3 \times 100\text{ K}} = 4.0 \times 10^{-4}\text{ K}^{-1}

Apparent expansivity is defined as the fraction of initial volume overflowed per degree rise in temperature.

3

Calculate the volume expansivity of the container

\gamma_v

\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Sum apparent expansivity and vessel cubic expansivity to find real cubic expansivity

\gamma_r

\gamma_r = \gamma_a + \gamma_v = 4.0 \times 10^{-4} + 0.6 \times 10^{-4} = 4.6 \times 10^{-4}\text{ K}^{-1}

Real expansivity accounts for both the apparent expansion of the liquid and the expansion of the container.

Anahtar Kavram

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel expansivity

Soru 1506Soru →

An X-ray tube operates at an accelerating potential difference of $25.0\text{ kV}$ . Calculate the maximum frequency of the emitted X-ray radiation in units of $10^{18}\text{ Hz}$ . (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ and elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ ).

Cevabı ve açıklamayı göster

Cevap: 6.03

Cevap

The maximum frequency of the emitted X-rays is

6.03 \times 10^{18}\text{ Hz}

, which gives a value of

6.03

in units of

10^{18}\text{ Hz}

.

The maximum frequency of X-ray photons emitted occurs when an accelerating electron transfers all of its kinetic energy (

e V

) into a single photon (

h f_{\text{max}}

). Substituting

e = 1.60 \times 10^{-19}\text{ C}

,

V = 25,000\text{ V}

, and

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

gives

f_{\text{max}} = 6.03 \times 10^{18}\text{ Hz}

, which equals

6.03

in the requested unit of

10^{18}\text{ Hz}

.

Adım Adım Çözüm

1

Convert potential difference from kilovolts to volts

V = 25.0\text{ kV} = 25,000\text{ V}

Standard SI units are required for calculations.

2

Determine maximum kinetic energy of the incident electrons

E_{\text{max}} = e V = 1.60 \times 10^{-19}\text{ C} \times 25,000\text{ V} = 4.00 \times 10^{-15}\text{ J}

The maximum photon energy produced equals the full kinetic energy acquired by an accelerated electron.

3

Calculate maximum frequency

f_{\text{max}}

using Duane-Hunt relation

f_{\text{max}} = \frac{E_{\text{max}}}{h} = \frac{4.00 \times 10^{-15}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.03 \times 10^{18}\text{ Hz}

According to the Duane-Hunt law,

e V = h f_{\text{max}}

.

Anahtar Kavram

Duane-Hunt Law and Maximum X-ray Frequency

Soru 1507Soru →

The threshold wavelength for photoelectric emission from a metallic surface is $500\text{ nm}$ . What is the work function of the metal in electron-volts ( $\text{eV}$ )? [Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^{8}\text{ m s}^{-1}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ ]

Cevabı ve açıklamayı göster

Cevap: 2.48

Cevap

The work function of the metal is

2.48\text{ eV}

(or

2.475\text{ eV}

).

The work function

W_0

is calculated using

W_0 = \frac{hc}{\lambda_0}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

,

c = 3.0 \times 10^8\text{ m s}^{-1}

, and

\lambda_0 = 500 \times 10^{-9}\text{ m}

gives

W_0 = 3.96 \times 10^{-19}\text{ J}

. Converting to electron-volts yields

\frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

, which rounds to

2.48\text{ eV}

.

Adım Adım Çözüm

1

State the relationship between work function and threshold wavelength

W_0 = \frac{hc}{\lambda_0}

The work function is the minimum energy required to liberate an electron, which corresponds to the maximum wavelength (threshold wavelength

\lambda_0

) that can cause emission.

2

Substitute the physical constants and threshold wavelength to calculate

W_0

in joules

W_0 = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{500 \times 10^{-9}} = 3.96 \times 10^{-19}\text{ J}

Evaluating

hc / \lambda_0

yields the work function energy in standard SI units (Joules).

3

Convert the calculated work function into electron-volts

W_0 = \frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

Dividing the energy in Joules by

1.6 \times 10^{-19}\text{ J/eV}

converts the value to electron-volts.

Anahtar Kavram

Work Function and Threshold Wavelength Relationship

Soru 1508Soru →

The dynamic pressure $P$ exerted by a moving fluid depends on its density $\rho$ and flow velocity $v$ according to the relationship $P = k \rho^a v^b$ , where $k$ is a dimensionless constant. What is the value of the exponent $b$ ?

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Cevap: 2

Cevap

The value of the exponent

b

is 2.

Equating the exponent of time (T) on both sides of the dimensional equation

\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}

yields

-2 = -b

, giving

b = 2

.

Adım Adım Çözüm

1

Determine the base dimensions of all physical quantities in the equation.

Pressure

[P] = \text{M L}^{-1} \text{T}^{-2}

, Density

[\rho] = \text{M L}^{-3}

, and Velocity

[v] = \text{L T}^{-1}

.

Dimensional analysis requires converting derived physical quantities into fundamental dimensions of Mass (M), Length (L), and Time (T).

2

Apply dimensional homogeneity by substituting the dimensions into

P = k \rho^a v^b

.

\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}

The principle of dimensional homogeneity states that exponents of M, L, and T must match on both sides of a physically correct equation.

3

Equate the corresponding exponents for time (T) and solve for

b

.

-2 = -b \implies b = 2

Matching the powers of T directly yields the numerical value of exponent

b

.

Anahtar Kavram

Principle of Dimensional Homogeneity and Dimensional Analysis

Soru 1509Soru →

A rectangular metallic sheet with a linear expansivity of $1.8 \times 10^{-5}\text{ K}^{-1}$ experiences a temperature rise of $50\text{ K}$ . If the increase in its surface area is $0.90\text{ cm}^2$ , what was the initial surface area of the sheet in $\text{cm}^2$ ?

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Cevap: 500

Cevap

The initial surface area of the metallic sheet is

500\text{ cm}^2

.

The initial area is found by converting linear expansivity to area expansivity (\beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}) and substituting into the area expansion relation \Delta A = A_0 \beta \Delta T, giving A_0 = \frac{0.90}{3.6 \times 10^{-5} \times 50} = 500\text{ cm}^2$.

Adım Adım Çözüm

1

Calculate the area (superficial) expansivity (\beta)

\beta = 2\alpha = 2 \times 1.8 \times 10^{-5}\text{ K}^{-1} = 3.6 \times 10^{-5}\text{ K}^{-1}

Surface area expansion depends on area expansivity, which is twice the linear expansivity for an isotropic solid.

2

Formulate the thermal area expansion equation

\Delta A = A_0 \beta \Delta T

The fractional change in area is directly proportional to the area expansivity and the temperature change.

3

Rearrange the formula to solve for the initial surface area (A_0)

A_0 = \frac{\Delta A}{\beta \Delta T}

Isolating the required unknown quantity.

4

Substitute the known numerical values and compute

A_0 = \frac{0.90\text{ cm}^2}{(3.6 \times 10^{-5}\text{ K}^{-1})(50\text{ K})} = \frac{0.90}{1.8 \times 10^{-3}} = 500\text{ cm}^2

Evaluating the expression yields the exact initial surface area.

Anahtar Kavram

Relationship between Linear Expansivity and Area Expansivity

Alternatif Yöntem

Calculate fractional area expansion per kelvin: \beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}

. Total fractional expansion for

50\text{ K}

is

3.6 \times 10^{-5} \times 50 = 0.0018

. Then initial area

A_0 = \frac{0.90}{0.0018} = 500\text{ cm}^2$.

Tahmini Süre:1m 30s

Soru 1510Soru →

At a coastal meteorological station in Calabar, air temperature observations recorded at four specific intervals during a 24-hour cycle were $22.0^\circ\text{C}$ , $31.5^\circ\text{C}$ , $28.5^\circ\text{C}$ , and $24.0^\circ\text{C}$ . What is the mean temperature of these observations in $^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 26.5

Cevap

The mean temperature computed from the four observations is

26.5^\circ\text{C}

.

The mean temperature is calculated by summing all recorded values (

22.0 + 31.5 + 28.5 + 24.0 = 106.0^\circ\text{C}

) and dividing by the total number of observations (

4

), resulting in

26.5^\circ\text{C}

.

Adım Adım Çözüm

1

Sum all four temperature values taken across the observation periods.

The total sum is

22.0^\circ\text{C} + 31.5^\circ\text{C} + 28.5^\circ\text{C} + 24.0^\circ\text{C} = 106.0^\circ\text{C}

.

Finding the arithmetic mean requires calculating the aggregate total of all recorded temperature data points.

2

Divide the calculated aggregate sum by the count of observation intervals.

\frac{106.0^\circ\text{C}}{4} = 26.5^\circ\text{C}

.

Dividing the sum by the sample size (

4

) yields the average temperature across the recorded intervals.

Anahtar Kavram

Calculation of mean temperature from periodic daily observations

Tahmini Süre:1m 30s

Soru 1511Soru →

A sealed cylinder in an industrial pneumatic lift contains $0.080\text{ m}^3$ of gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The piston compresses the gas to a final volume of $0.020\text{ m}^3$ , raising the pressure to $7.50 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius ( $^\circ\text{C}$ )?

Cevabı ve açıklamayı göster

Cevap: 102

Cevap

The final temperature of the gas is

102^\circ\text{C}

.

Using the combined gas law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with absolute initial temperature

T_1 = 300\text{ K}

gives

T_2 = 375\text{ K}

. Subtracting

273

yields the final temperature of

102^\circ\text{C}

.

Adım Adım Çözüm

1

Convert initial temperature to absolute temperature (Kelvin)

T_1 = 27 + 273 = 300\text{ K}

Gas laws require thermodynamic temperature measured on the Kelvin scale.

2

Set up the Combined Gas Law equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final states when pressure, volume, and temperature all change for a fixed mass of gas.

3

Calculate the final absolute temperature

T_2

T_2 = 300 \times \frac{7.50 \times 10^5 \times 0.020}{1.50 \times 10^5 \times 0.080} = 375\text{ K}

Evaluating the ratio of

P_2 V_2

to

P_1 V_1

yields

1.25

.

4

Convert final temperature from Kelvin to Celsius

\theta_2 = 375 - 273 = 102^\circ\text{C}

The question explicitly requires the answer in degrees Celsius.

Anahtar Kavram

Combined Gas Law and thermodynamic temperature conversion

Soru 1512Soru →

An aluminum electric cable suspended between two transmission poles has a length of $100\text{ m}$ at an initial morning temperature of $20^\circ\text{C}$ . By afternoon, the cable temperature increases to $50^\circ\text{C}$ . Given that the linear expansivity of aluminum is $2.3 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the length of the cable in centimeters?

Cevabı ve açıklamayı göster

Cevap: 6.9

Cevap

The increase in the length of the cable is

6.9\text{ cm}

.

Applying the formula for linear expansion

\Delta L = L_0 \alpha \Delta T

, where

L_0 = 100\text{ m}

,

\alpha = 2.3 \times 10^{-5}\text{ K}^{-1}

, and

\Delta T = 30\text{ K}

, gives

\Delta L = 0.069\text{ m}

. Converting this to centimeters yields

6.9\text{ cm}

.

Adım Adım Çözüm

1

Determine the temperature change

\Delta T = 30\text{ K}

Temperature change is the difference between final and initial temperatures:

50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}

.

2

Calculate expansion in meters using the linear expansivity formula

\Delta L = 0.069\text{ m}

\Delta L = L_0 \alpha \Delta T = 100 \times (2.3 \times 10^{-5}) \times 30 = 0.069\text{ m}.

3

Convert the change in length to centimeters

\Delta L = 6.9\text{ cm}

Since

1\text{ m} = 100\text{ cm}

, multiply

0.069\text{ m}

by

100

to obtain

6.9\text{ cm}

.

Anahtar Kavram

Linear expansivity of solid conductors

Tahmini Süre:1m 30s

Soru 1513Soru →

A glass flask of volume $800\text{ cm}^3$ at $10^\circ\text{C}$ is completely filled with a liquid having a real volume expansivity of $5.0 \times 10^{-4}\text{ K}^{-1}$ . If the linear expansivity of the glass is $1.0 \times 10^{-5}\text{ K}^{-1}$ , what volume of the liquid (in $\text{cm}^3$ ) will overflow when the flask and its contents are heated to $60^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 18.8

Cevap

18.8 cm^3

The volume of liquid that overflows is equal to the apparent increase in volume of the liquid. The apparent volume expansivity of the liquid $\gamma_a$ is given by $\gamma_a = \gamma_r - \gamma_v$, where $\gamma_r = 5.0 \times 10^{-4}\text{ K}^{-1}$ is the real expansivity of the liquid, and $\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5} = 0.3 \times 10^{-4}\text{ K}^{-1}$ is the volume expansivity of the glass flask. Therefore, $\gamma_a = 5.0 \times 10^{-4} - 0.3 \times 10^{-4} = 4.7 \times 10^{-4}\text{ K}^{-1}$. The overflow volume is $\Delta V_a = V_0 \times \gamma_a \times \Delta T = 800 \times (4.7 \times 10^{-4}) \times 50 = 18.8\text{ cm}^3$.

Adım Adım Çözüm

1

Calculate the cubic expansivity of the glass flask

$\gamma_v = 3.0 \times 10^{-5}\text{ K}^{-1} = 0.3 \times 10^{-4}\text{ K}^{-1}$

The volume expansivity of a solid container is three times its linear expansivity ($\gamma_v = 3\alpha$).

2

Determine the apparent volume expansivity of the liquid

$\gamma_a = 4.7 \times 10^{-4}\text{ K}^{-1}$

Apparent expansivity is equal to real expansivity minus the vessel cubic expansivity ($\gamma_a = \gamma_r - \gamma_v$).

3

Calculate the change in temperature

$\Delta T = 50\text{ K}$

The temperature increases from

10^\circ\text{C}

to

60^\circ\text{C}

.

4

Calculate the overflow volume

$\Delta V_a = 18.8\text{ cm}^3$

The overflow volume is the apparent expansion of the liquid given by $\Delta V_a = V_0 \gamma_a \Delta T$.

Anahtar Kavram

Real and apparent cubic expansivity of liquids

Soru 1514Soru →

A geography student constructs a pie chart to illustrate the annual export earnings of agricultural commodities from a region. The total value of all exports is $\text{\$}480\text{ million}$ , with oil palm products accounting for $\text{\$}108\text{ million}$ . What is the central angle, in degrees, of the sector representing oil palm products?

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Cevap: 81

Cevap

The central angle of the sector representing oil palm products is

81^\circ

.

The correct answer is obtained by finding the component share of the total value (108 / 480 = 0.225) and multiplying by 360 degrees, yielding an angle of 81 degrees.

Adım Adım Çözüm

1

Calculate the fractional proportion of the specific component relative to the total data value

\frac{108}{480} = 0.225

A sector in a pie chart represents a fractional part of the total statistical value.

2

Convert the calculated proportion into degrees of a circle

0.225 \times 360^\circ = 81^\circ

A full pie chart circle contains

360^\circ

, so multiplying the component proportion by

360^\circ

gives its angular representation.

Anahtar Kavram

Calculating Sector Angles in Pie Charts

Soru 1515Soru →

A cell of electromotive force $E$ and internal resistance $r$ supplies a current of $1.2\text{ A}$ when connected across a $4.0\ \Omega$ resistor. When this resistor is replaced with a $10.0\ \Omega$ resistor, the current decreases to $0.6\text{ A}$ . Calculate the internal resistance $r$ of the cell in ohms ( $\Omega$ ).

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Cevap: 2

Cevap

The internal resistance of the cell is

2.0\ \Omega

.

For a complete circuit, the electromotive force

E

is related to load resistance

R

, internal resistance

r

, and current

I

by

E = I(R + r)

. Setting up equations for both load conditions gives

E = 1.2(4.0 + r)

and

E = 0.6(10.0 + r)

. Equating these expressions yields

4.8 + 1.2r = 6.0 + 0.6r

, which simplifies to

0.6r = 1.2

, giving an internal resistance of

r = 2.0\ \Omega

.

Adım Adım Çözüm

1

Formulate the circuit equation for the first load resistance

E = 1.2 \times (4.0 + r) = 4.8 + 1.2r

Using Ohm's law for a complete circuit, the electromotive force equals current times total circuit resistance:

E = I(R_1 + r)

.

2

Formulate the circuit equation for the second load resistance

E = 0.6 \times (10.0 + r) = 6.0 + 0.6r

The cell maintains the same internal electromotive force

E

and internal resistance

r

with the new load resistor

R_2

.

3

Equate the expressions for

E

and solve for

r

4.8 + 1.2r = 6.0 + 0.6r \implies 0.6r = 1.2 \implies r = 2.0\ \Omega

Since

E

is constant for the cell, setting the two right-hand sides equal yields a single linear equation for the unknown internal resistance

r

.

Anahtar Kavram

Electromotive Force and Internal Resistance

Soru 1516Soru →

An object is launched vertically upward from the ground with an initial velocity of $40\text{ m/s}$ . At the exact same instant, a second object is dropped from rest from a height of $100\text{ m}$ directly above the first object. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , at what time (in seconds) after launch will the two objects meet?

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Cevap: 2.5

Cevap

The two objects meet after 2.5 seconds.

Because both objects experience identical downward gravitational acceleration (

g = 10\text{ m/s}^2

), their relative acceleration is zero. The relative velocity between them remains constant at

40\text{ m/s}

. The time to cover the initial separation distance of

100\text{ m}

is calculated directly as

t = \frac{\text{distance}}{\text{relative velocity}} = \frac{100}{40} = 2.5\text{ s}

.

Adım Adım Çözüm

1

Set up the position-time equations for both objects taking ground level as

y = 0\text{ m}

.

For the upward-launched object:

y_1(t) = u t - \frac{1}{2}gt^2 = 40t - 5t^2

. For the dropped object:

y_2(t) = h_0 - \frac{1}{2}gt^2 = 100 - 5t^2

.

Kinematic equations of motion under uniform gravitational acceleration apply to both bodies.

2

Equate the two vertical position equations to solve for the meeting time

t

.

40t - 5t^2 = 100 - 5t^2 \implies 40t = 100

When the objects meet, they share the exact same vertical position

y_1(t) = y_2(t)

at time

t

.

3

Calculate the value of

t

.

t = \frac{100}{40} = 2.5\text{ s}

Direct algebraic division yields the time elapsed before collision/meeting.

Anahtar Kavram

Relative vertical motion under uniform gravity

Tahmini Süre:1m 30s

Soru 1517Soru →

In an industrial non-destructive testing setup, an X-ray tube produces continuous X-radiation with a minimum cut-off wavelength of $3.3 \times 10^{-11}\text{ m}$ . What is the operating accelerating potential difference of the tube in kilovolts ( $\text{kV}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^8\text{ m s}^{-1}$ , and electron charge $e = 1.6 \times 10^{-19}\text{ C}$ ).

Cevabı ve açıklamayı göster

Cevap: 37.5

Cevap

The operating accelerating potential difference of the tube is 37.5 kV.

According to the Duane-Hunt law, the maximum kinetic energy of electrons hitting the target equal the maximum photon energy produced:

e V = h f_{\text{max}} = \frac{h c}{\lambda_{\text{min}}}

. Rearranging to solve for voltage gives

V = \frac{h c}{e \lambda_{\text{min}}}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

,

c = 3.0 \times 10^8\text{ m s}^{-1}

,

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda_{\text{min}} = 3.3 \times 10^{-11}\text{ m}

yields

V = 37,500\text{ V}

, which equals

37.5\text{ kV}

.

Adım Adım Çözüm

1

Relate the maximum electron kinetic energy to the shortest X-ray photon wavelength using Duane-Hunt law

e V = E_{\text{max}} = \frac{h c}{\lambda_{\text{min}}}

At the Duane-Hunt cutoff limit, the entire kinetic energy of an accelerating electron is converted into a single X-ray photon.

2

Rearrange the equation to solve for the accelerating voltage

V

V = \frac{h c}{e \lambda_{\text{min}}}

Isolating

V

allows direct calculation from known fundamental constants and the given minimum wavelength.

3

Substitute the physical constants and calculate the value of

V

in Volts

V = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m s}^{-1})}{(1.6 \times 10^{-19}\text{ C})(3.3 \times 10^{-11}\text{ m})} = 37,500\text{ V}

Carrying out arithmetic with scientific notation powers yields

3.75 \times 10^4\text{ V}

.

4

Convert the potential difference from Volts (V) to kilovolts (kV)

37,500\text{ V} = 37.5\text{ kV}

Dividing by 1000 converts potential difference into the requested kilovolt unit.

Anahtar Kavram

Duane-Hunt Law and Cut-off Wavelength in X-ray Production

Soru 1518Soru →

An atom has three stationary energy levels given by $E_1 = -12.50\text{ eV}$ , $E_2 = -6.80\text{ eV}$ , and $E_3 = -3.50\text{ eV}$ . What is the wavelength, in nanometers ( $\text{nm}$ ), of the photon emitted during the transition that produces the longest wavelength line in its emission spectrum? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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Cevap: 375

Cevap

The wavelength of the photon emitted for the longest wavelength spectral line is 375 nm.

Photon wavelength is related to transition energy by

\lambda = \frac{hc}{\Delta E}

. To find the longest wavelength spectral line, the transition with the smallest energy gap must be used. Evaluating all emission transitions between the levels gives

\Delta E_{3 \to 2} = 3.30\text{ eV}

,

\Delta E_{2 \to 1} = 5.70\text{ eV}

, and

\Delta E_{3 \to 1} = 9.00\text{ eV}

. The minimum energy difference is

3.30\text{ eV}

. Converting

3.30\text{ eV}

to Joules gives

3.30 \times 1.6 \times 10^{-19} = 5.28 \times 10^{-19}\text{ J}

. Substituting this into the wavelength formula yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{5.28 \times 10^{-19}} = 3.75 \times 10^{-7}\text{ m} = 375\text{ nm}

.

Adım Adım Çözüm

1

Determine which electronic transition yields the longest wavelength photon.

Transition from

E_3

to

E_2

yields the minimum energy difference of

3.30\text{ eV}

.

Since

\lambda = \frac{hc}{\Delta E}

, the longest wavelength corresponds to the smallest energy transition.

2

Convert the transition energy from electron-volts to Joules.

\Delta E = 5.28 \times 10^{-19}\text{ J}

.

SI units are required for calculations involving Planck's constant and the speed of light.

3

Calculate the wavelength

\lambda

using the photon energy formula

\lambda = \frac{hc}{\Delta E}

.

\lambda = 375\text{ nm}

.

Substituting

h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}

,

c = 3.0 \times 10^8\text{ m/s}

, and

\Delta E = 5.28 \times 10^{-19}\text{ J}

gives

3.75 \times 10^{-7}\text{ m}

, which equals

375\text{ nm}

.

Anahtar Kavram

Inverse relationship between transition energy and photon wavelength in atomic emission spectra

Soru 1519Soru →

A sand bag of mass $8.0\text{ kg}$ is suspended vertically by a light rope. A projectile of mass $0.50\text{ kg}$ moving horizontally at a speed of $170\text{ m s}^{-1}$ strikes the sand bag and becomes embedded in it. What is the common speed, in $\text{m s}^{-1}$ , of the sand bag and the embedded projectile immediately after collision?

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Cevap: 10

Cevap

The common speed of the sand bag and embedded projectile immediately after the collision is

10.0\text{ m s}^{-1}

.

According to the principle of conservation of linear momentum, total momentum before impact equals total momentum after impact. The initial momentum of the system is entirely from the projectile:

p_i = 0.50 \times 170 = 85\text{ kg m s}^{-1}

. After collision, both objects move together with a total mass of

0.50 + 8.0 = 8.5\text{ kg}

. Setting

8.5 v = 85

gives

v = 10\text{ m s}^{-1}

.

Adım Adım Çözüm

1

State the conservation of linear momentum equation for an inelastic collision

m_1 u_1 + m_2 u_2 = (m_1 + m_2) v

Since no net external horizontal force acts on the system during impact, total linear momentum is conserved.

2

Substitute the given values into the momentum balance equation

(0.50\text{ kg})(170\text{ m s}^{-1}) + (8.0\text{ kg})(0\text{ m s}^{-1}) = (0.50\text{ kg} + 8.0\text{ kg}) v

The projectile embeds into the sand bag, so they move together with a single combined mass.

3

Solve the linear equation for the common final velocity

v

85 = 8.5 v \implies v = 10\text{ m s}^{-1}

Dividing total initial momentum by total combined mass yields the final speed.

Anahtar Kavram

Conservation of Linear Momentum in Completely Inelastic Collisions

Tahmini Süre:1m 30s

Soru 1520Soru →

A particle traveling along a straight line accelerates uniformly from an initial speed $u$ to a final speed of $25\text{ m/s}$ over a distance of $150\text{ m}$ in a time duration of $8\text{ s}$ . What is the initial speed $u$ in $\text{m/s}$ ?

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Cevap: 12.5

Cevap

The initial speed of the particle is

12.5\text{ m/s}

.

Under uniform acceleration, displacement is given by the product of average velocity and time:

s = \frac{u + v}{2}t

. Substituting

s = 150\text{ m}

,

v = 25\text{ m/s}

, and

t = 8\text{ s}

gives

150 = 4(u + 25)

, which yields

u = 12.5\text{ m/s}

.

Adım Adım Çözüm

1

Identify the given parameters and select the appropriate kinematic equation.

Givens:

s = 150\text{ m}

,

v = 25\text{ m/s}

,

t = 8\text{ s}

. Formula:

s = \frac{u + v}{2}t

.

This formula relates displacement directly to average velocity and time under uniform acceleration without needing the acceleration variable.

2

Substitute the given values into the formula.

150 = \left(\frac{u + 25}{2}\right) \times 8 = 4(u + 25)

.

Simplifying

\frac{8}{2}

gives a factor of

4

multiplying

(u + 25)

.

3

Isolate and solve for the unknown initial speed

u

.

u + 25 = \frac{150}{4} = 37.5 \implies u = 12.5\text{ m/s}

.

Subtracting

25

from

37.5

yields the value of

u

.

Anahtar Kavram

Kinematics with uniform linear acceleration

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