Tüm alıştırma soruları

Soru 841Soru →

A $5.0\text{ }\mu\text{F}$ parallel-plate capacitor is connected across a potential difference of $20.0\text{ V}$ . What is the magnitude of the electric charge stored on either plate of the capacitor, in microcoulombs ( $\mu\text{C}$ )?

Cevabı ve açıklamayı göster

Cevap: 100

Cevap

The magnitude of the electric charge stored on either plate of the capacitor is

100.0\text{ }\mu\text{C}

.

Using the capacitor charge equation

Q = C \times V

, substituting

C = 5.0\text{ }\mu\text{F}

and

V = 20.0\text{ V}

yields

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Adım Adım Çözüm

1

Identify the given physical quantities and the formula for electric charge on a capacitor.

Capacitance

C = 5.0\text{ }\mu\text{F}

, voltage

V = 20.0\text{ V}

. Formula:

Q = C V

.

The charge stored by a capacitor is directly proportional to the potential difference across its terminals.

2

Perform the multiplication to determine the charge magnitude in microcoulombs.

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Multiplying capacitance in microfarads by potential difference in volts yields charge directly in microcoulombs.

Anahtar Kavram

Fundamental relationship between capacitance, charge, and potential difference (

Q = C V

)

Tahmini Süre:45s

Soru 842Soru →

A motorcycle traveling at a constant speed of $18\text{ m/s}$ passes a landmark. Exactly $4.0\text{ s}$ after passing the landmark, the rider accelerates uniformly at a rate of $2.5\text{ m/s}^2$ until reaching a speed of $28\text{ m/s}$ . What is the total distance, in meters, traveled by the motorcycle from the instant it passed the landmark to the moment it reaches $28\text{ m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 164

Cevap

The total distance traveled by the motorcycle is

164\text{ m}

.

The total distance is obtained by finding the displacement during the

4.0\text{ s}

of constant speed at

18\text{ m/s}

(

72\text{ m}

) and adding the displacement during uniform acceleration from

18\text{ m/s}

to

28\text{ m/s}

at

2.5\text{ m/s}^2

(

92\text{ m}

), giving a sum of

164\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance covered during the initial constant speed stage.

The distance covered in the first

4.0\text{ s}

is

72\text{ m}

.

At a constant velocity

v = 18\text{ m/s}

, the distance

s_1 = v \times t = 18 \times 4.0 = 72\text{ m}

.

2

Calculate the distance covered during the accelerated motion stage.

The distance covered while accelerating from

18\text{ m/s}

to

28\text{ m/s}

is

92\text{ m}

.

Using the kinematic equation

v^2 = u^2 + 2as_2

, substitute

u = 18\text{ m/s}

,

v = 28\text{ m/s}

, and

a = 2.5\text{ m/s}^2

to get

28^2 = 18^2 + 2(2.5)s_2 \implies 784 = 324 + 5s_2 \implies s_2 = 92\text{ m}

.

3

Find the total distance traveled.

Total distance

s_{\text{total}} = 164\text{ m}

.

The total distance is the sum of the distance covered during the constant speed period (

72\text{ m}

) and during uniform acceleration (

92\text{ m}

).

Anahtar Kavram

Kinematics equations for multi-stage motion combining uniform speed and uniform acceleration.

Soru 843Soru →

A solid metal block of mass $4.0\text{ kg}$ has a density of $8000\text{ kg m}^{-3}$ at $20^\circ\text{C}$ . If the linear expansivity of the metal is $2.0 \times 10^{-5}\text{ K}^{-1}$ , calculate the increase in volume of the block, in $\text{cm}^3$ , when its temperature is raised to $120^\circ\text{C}$ .

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The increase in volume of the block is

3.0\text{ cm}^3

.

First, the initial volume of the metal block is found using

V_0 = \frac{m}{\rho} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3

. Next, because a solid expands in three dimensions, the cubic expansivity is

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

. The temperature change is

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

. Finally, the increase in volume is computed as

\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3

.

Adım Adım Çözüm

1

Calculate the initial volume of the metal block from its mass and density.

V_0 = \frac{m}{\rho_0} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3

Volume is equal to mass divided by density.

2

Convert linear expansivity (

\alpha

) to volume expansivity (

\gamma

).

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

For an isotropic solid, cubic expansivity is three times its linear expansivity.

3

Determine the change in temperature.

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

Thermal expansion is driven by the change in temperature.

4

Calculate the total volume expansion.

\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3

Applying the thermal volume expansion formula

\Delta V = V_0 \gamma \Delta T

.

Anahtar Kavram

Thermal Volume Expansion of Solids

Tahmini Süre:2m 0s

Soru 844Soru →

The cumulative frequency distribution below shows the monthly electricity consumption (in kWh) recorded for $50$ households in a residential estate:

Electricity Usage (kWh)	Cumulative Frequency
$\leq 50$	5
$\leq 100$	15
$\leq 150$	35
$\leq 200$	45
$\leq 250$	50

Using linear interpolation, calculate the 60th percentile ( $P_{60}$ ) of electricity consumption in kWh.

Cevabı ve açıklamayı göster

Cevap: 137.5

Cevap

137.5 kWh

The 60th percentile corresponds to the 30th observation out of 50 (

0.60 \times 50 = 30

). From the cumulative frequency table, the 30th value falls in the class interval between

100

and

150

. Using the lower boundary of

100

, previous cumulative frequency of

15

, interval frequency of

20

, and interval width of

50

, linear interpolation gives

100 + \frac{30 - 15}{20} \times 50 = 137.5

kWh.

Adım Adım Çözüm

1

Determine the rank for the 60th percentile

Rank position = 30th household

The 60th percentile rank corresponds to

60\%

of the total sample size (

N = 50

), which is

0.60 \times 50 = 30

.

2

Identify the parameters of the percentile class interval

Lower boundary

L = 100

,

CF_{\text{prev}} = 15

, class frequency

f = 20

, width

w = 50

The cumulative frequency rises from 15 to 35 in the class interval

(100, 150]

, so the 30th value lies within this interval.

3

Calculate the value using linear interpolation

137.5 kWh

Substitute values into

P_{60} = 100 + \left(\frac{30 - 15}{20}\right) \times 50 = 100 + 37.5 = 137.5

kWh.

Anahtar Kavram

Linear interpolation for percentiles from a cumulative frequency distribution

Soru 845Soru →

Find the real value of $x$ that satisfies the logarithmic equation $\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The real value of

x

that satisfies the equation is

6

.

Applying the logarithmic quotient rule transforms

\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3

into

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

. Expressing this in exponential form yields

\frac{x^2 + 3x - 22}{x - 2} = 8

, which simplifies to

x^2 - 5x - 6 = 0

. Factoring gives

(x - 6)(x + 1) = 0

. Since the argument of a logarithm must be strictly positive (

x - 2 > 0 \implies x > 2

),

x = -1

is extraneous and

x = 6

is the only valid solution.

Adım Adım Çözüm

1

Combine the logarithmic terms using the quotient law of logarithms.

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

The difference of two logarithms with the same base equals the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Convert the logarithmic equation into its equivalent exponential form.

\frac{x^2 + 3x - 22}{x - 2} = 2^3 = 8

If

\log_b Y = c

, then

Y = b^c

.

3

Clear the denominator and simplify to form a quadratic equation.

x^2 - 5x - 6 = 0

Multiplying both sides by

(x - 2)

yields

x^2 + 3x - 22 = 8x - 16

, which rearranges to

x^2 - 5x - 6 = 0

.

4

Solve the quadratic equation by factoring.

(x - 6)(x + 1) = 0 \implies x = 6 \text{ or } x = -1

The factors of

-6

that sum to

-5

are

-6

and

1

.

5

Test roots against domain restrictions to eliminate extraneous solutions.

x = 6

Logarithmic arguments must be strictly positive. For

\log_2(x - 2)

to be defined,

x > 2

. Thus,

x = -1

is extraneous, leaving

x = 6

as the unique valid solution.

Anahtar Kavram

Logarithmic Equations and Domain Restrictions

Soru 846Soru →

A straight line $L$ passes through the point $(2, -3)$ and is perpendicular to the line $4x + 5y - 20 = 0$ . Calculate the $x$ -intercept of line $L$ .

Cevabı ve açıklamayı göster

Cevap: 4.4

Cevap

The

x

-intercept of line

L

is

4.4

(or

\frac{22}{5}

).

The given line

4x + 5y - 20 = 0

has a slope of

-\frac{4}{5}

. The line perpendicular to it must have a slope equal to the negative reciprocal, which is

\frac{5}{4}

. Using the point-slope formula with point

(2, -3)

, the equation of line

L

is

y + 3 = \frac{5}{4}(x - 2)

, simplifying to

y = \frac{5}{4}x - \frac{11}{2}

. Setting

y = 0

yields

\frac{5}{4}x = \frac{11}{2}

, giving an

x

-intercept of

x = \frac{22}{5} = 4.4

.

Adım Adım Çözüm

1

Determine the gradient of the given line

4x + 5y - 20 = 0

Gradient

m_1 = -\frac{4}{5}

Converting to slope-intercept form

y = -\frac{4}{5}x + 4

reveals the slope.

2

Calculate the perpendicular gradient for line

L

Gradient

m = \frac{5}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Derive the equation of line

L

using point

(2, -3)

y = \frac{5}{4}x - \frac{11}{2}

Substitute the point

(2, -3)

and gradient

m = \frac{5}{4}

into point-slope form.

4

Solve for the

x

-intercept by setting

y = 0

x = 4.4

The

x

-intercept is defined as the point where the line crosses the

x

-axis (

y = 0

).

Anahtar Kavram

Perpendicular line slope relationships and x-intercept calculations

Tahmini Süre:1m 30s

Soru 847Soru →

A mechanical wave propagating along a string is defined by the displacement equation $y = 0.05 \sin\left(160\pi t - 8\pi x + \frac{\pi}{3}\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. The wave transitions into a different section of string where its propagation speed drops by $25\%$ . Calculate the minimum distance (in meters) between two points in this second section that have a phase difference of $\frac{2\pi}{3}\text{ rad}$ .

Cevabı ve açıklamayı göster

Cevap: 0.0625

Cevap

The minimum distance between the two points in the second section is

0.0625\text{ m}

.

The correct calculation gives

0.0625\text{ m}

. Comparing

y = 0.05 \sin(160\pi t - 8\pi x + \pi/3)

with the general form

y = A \sin(\omega t - k x + \phi_0)

identifies

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

, yielding an initial speed of

v_1 = \omega / k_1 = 20\text{ m/s}

. Upon transitioning into the second string section, the speed drops by

25\%

to

v_2 = 15\text{ m/s}

. Since the angular frequency

\omega = 160\pi\text{ rad/s}

remains invariant during refraction, the wave number in the second section is

k_2 = \omega / v_2 = 160\pi / 15 = 32\pi / 3\text{ rad/m}

. Substituting

k_2

and the given phase difference

\Delta \phi = 2\pi / 3\text{ rad}

into

\Delta \phi = k_2 \Delta x

yields

\Delta x = (2\pi / 3) / (32\pi / 3) = 2/32 = 0.0625\text{ m}

.

Adım Adım Çözüm

1

Extract angular frequency and wave number from the displacement equation

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

The standard progressive wave equation is formatted as

y = A \sin(\omega t - k x + \phi_0)

.

2

Calculate the initial wave propagation speed

v_1

v_1 = \frac{\omega}{k_1} = \frac{160\pi}{8\pi} = 20\text{ m/s}

Wave speed is equal to the ratio of angular frequency to wave number.

3

Determine the wave speed

v_2

in the second section

v_2 = 20 \times (1 - 0.25) = 15\text{ m/s}

The wave speed decreases by

25\%

, making

v_2 = 0.75 v_1

.

4

Find the new wave number

k_2

in the second section

k_2 = \frac{\omega}{v_2} = \frac{160\pi}{15} = \frac{32\pi}{3}\text{ rad/m}

Frequency and angular frequency remain invariant when a wave passes from one medium to another.

5

Calculate the spatial separation

\Delta x

for the specified phase difference

\Delta x = \frac{\Delta \phi}{k_2} = \frac{2\pi / 3}{32\pi / 3} = \frac{2}{32} = 0.0625\text{ m}

Phase difference relates to spatial distance via

\Delta \phi = k \Delta x

.

Anahtar Kavram

Wave Equation Parameter Extraction and Invariance of Frequency in Refraction

Soru 848Soru →

A metal container with an initial volume of $500\text{ cm}^3$ is completely filled with an organic liquid at $15^\circ\text{C}$ . The real cubic expansivity of the liquid is $4.0 \times 10^{-4}\text{ K}^{-1}$ . When the container and liquid are heated together to a final temperature of $65^\circ\text{C}$ , exactly $8.5\text{ cm}^3$ of the liquid overflows. What is the linear expansivity of the metal container, in units of $10^{-5}\text{ K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The linear expansivity of the metal container is

2.0 \times 10^{-5}\text{ K}^{-1}

, which corresponds to a numerical value of

2.0

in units of

10^{-5}\text{ K}^{-1}

.

The apparent expansivity of the liquid is derived from the overflow volume (8.5 cm^3 / (500 cm^3 * 50 K) = 3.4 x 10^-4 K^-1). Subtracting this from the liquid's real expansivity (4.0 x 10^-4 K^-1) yields the container's cubic expansivity of 0.6 x 10^-4 K^-1 (or 6.0 x 10^-5 K^-1). Dividing by 3 gives the linear expansivity of the container material as 2.0 x 10^-5 K^-1.

Adım Adım Çözüm

1

Determine the temperature increase

\Delta T = 50\text{ K}

Temperature change drives the thermal expansion process.

2

Compute the apparent cubic expansivity of the liquid

\gamma_a = 3.4 \times 10^{-4}\text{ K}^{-1}

Apparent expansion is measured directly from the liquid overflow relative to initial volume and temperature rise.

3

Calculate the cubic expansivity of the metal container

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The difference between real cubic expansivity of the liquid and its apparent cubic expansivity equals the cubic expansivity of the container.

4

Calculate the linear expansivity of the container material

\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}

For isotropic solids, volume expansivity is three times linear expansivity (\gamma_v = 3\alpha).

Anahtar Kavram

Thermal expansion of liquids in expanding vessels: Real expansivity equals apparent expansivity plus vessel cubic expansivity (\gamma_r = \gamma_a + 3\alpha).

Tahmini Süre:2m 0s

Soru 849Soru →

A progressive transverse wave traveling along a medium is described by the displacement equation $y = 0.05 \sin\left(160\pi t - 4\pi x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the ratio of the maximum particle velocity to the wave propagation velocity? (Take $\pi = 3.142$ .)

Cevabı ve açıklamayı göster

Cevap: 0.628

Cevap

The ratio of the maximum particle velocity to the wave propagation velocity is 0.628.

For a progressive wave

y = A \sin(\omega t - k x)

, the wave advances at speed

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

. Meanwhile, individual particles vibrate transversely with simple harmonic motion where maximum velocity is

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s}

. Taking the ratio gives

\frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

, which rounds to 0.628.

Adım Adım Çözüm

1

Extract parameters from the standard wave equation

y = A \sin(\omega t - k x)

A = 0.05\text{ m}

,

\omega = 160\pi\text{ rad/s}

,

k = 4\pi\text{ rad/m}

Matching the given equation with standard progressive wave form yields the required wave constants.

2

Calculate the wave propagation velocity

v

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

The speed at which the wave energy advances through the medium depends on angular frequency and wave number.

3

Calculate the maximum transverse particle velocity

v_{p,\text{max}}

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s} \approx 25.136\text{ m/s}

Particles perform simple harmonic motion, whose maximum speed is given by the product of amplitude and angular frequency.

4

Compute the ratio of maximum particle velocity to wave velocity

Ratio

= \frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

Dividing the maximum particle speed by the wave speed gives the desired dimensionless ratio.

Anahtar Kavram

Distinction between particle oscillation velocity and wave propagation velocity

Soru 850Soru →

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from two fixed points $A(-2, 3)$ and $B(4, 1)$ . If the locus of $P$ intersects the horizontal line $y = 5$ at the point $(k, 5)$ , what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

k

is

2

.

The locus of points equidistant from

A(-2, 3)

and

B(4, 1)

is the perpendicular bisector of

AB

. The midpoint of

AB

is

(1, 2)

and the slope of

AB

is

-\frac{1}{3}

, giving a perpendicular slope of

3

. The equation of the locus is

3x - y = 1

. Substituting

y = 5

gives

3k - 5 = 1

, which yields

k = 2

.

Adım Adım Çözüm

1

Identify the nature of the locus

The locus of points equidistant from two fixed points

A

and

B

is the perpendicular bisector of the line segment

AB

.

By definition, the set of points equidistant from two fixed points forms a straight line perpendicular to the segment joining the two points at its midpoint.

2

Find the midpoint of segment

AB

Midpoint

M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2)

.

The perpendicular bisector passes through the midpoint of the line segment.

3

Calculate the gradient of

AB

and the perpendicular gradient

Gradient of

AB = \frac{1 - 3}{4 - (-2)} = -\frac{1}{3}

. Thus, the perpendicular gradient is

3

.

Perpendicular lines have gradients whose product is

-1

.

4

Derive the equation of the locus

y - 2 = 3(x - 1) \implies y = 3x - 1 \implies 3x - y = 1

.

Use the point-slope form of a line equation with point

(1, 2)

and slope

3

.

5

Determine the value of

k

at

y = 5

3k - 5 = 1 \implies 3k = 6 \implies k = 2

.

Substitute the point

(k, 5)

into the locus equation.

Anahtar Kavram

Perpendicular Bisector as a Locus

Tahmini Süre:1m 30s

Soru 851Soru →

Two identical isolated metal spheres carrying charges of $+8.0 \times 10^{-6}\text{ C}$ and $-2.0 \times 10^{-6}\text{ C}$ are brought into contact and then separated to a distance of $0.30\text{ m}$ in a vacuum. What is the magnitude of the electrostatic force of repulsion, in newtons ( $\text{N}$ ), between the spheres after contact? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{C}^{-2}$ )

Cevabı ve açıklamayı göster

Cevap: 0.9

Cevap

The magnitude of the electrostatic force of repulsion between the spheres after contact is

0.9\text{ N}

.

When identical conducting spheres touch, their total electric charge is conserved and shared equally. The net charge is

+8.0\,\mu\text{C} + (-2.0\,\mu\text{C}) = +6.0\,\mu\text{C}

, giving each sphere a charge of

+3.0\,\mu\text{C}

. Applying Coulomb's law with a distance of

0.30\text{ m}

yields

0.9\text{ N}

.

Adım Adım Çözüm

1

Calculate the net combined charge of the two identical spheres when brought into contact.

Q_{\text{total}} = q_1 + q_2 = (+8.0 \times 10^{-6}\text{ C}) + (-2.0 \times 10^{-6}\text{ C}) = +6.0 \times 10^{-6}\text{ C}

.

According to the principle of conservation of charge, charges add algebraically.

2

Determine the charge on each individual sphere after separation.

q' = \frac{Q_{\text{total}}}{2} = \frac{+6.0 \times 10^{-6}\text{ C}}{2} = +3.0 \times 10^{-6}\text{ C}

.

Identical conducting spheres share total charge equally when in contact.

3

Compute the force of repulsion using Coulomb's law.

F = \frac{k(q')^2}{r^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})^2}{(0.30)^2} = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-12}}{0.09} = 0.9\text{ N}

.

Coulomb's law defines the electrostatic force between two point charges.

Anahtar Kavram

Charge conservation, redistribution by conduction, and Coulomb's law

Soru 852Soru →

An electron inside an excited atom drops from an energy state of $-1.20\text{ eV}$ to a lower energy state of $-4.50\text{ eV}$ . What is the frequency of the emitted electromagnetic radiation, in units of $10^{14}\text{ Hz}$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The frequency of the emitted electromagnetic radiation is

8.0 \times 10^{14}\text{ Hz}

(or 8.0 in units of

10^{14}\text{ Hz}

).

The energy of the photon emitted during a downward transition between discrete atomic energy levels is equal to the energy difference between the initial and final states. Calculating

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

, converting to Joules gives

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

. Dividing this energy by Planck's constant

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

yields a frequency of

8.0 \times 10^{14}\text{ Hz}

.

Adım Adım Çözüm

1

Determine the energy of the emitted photon in electron-volts

\Delta E = 3.30\text{ eV}

The energy of the emitted photon equals the difference between the upper and lower atomic energy levels:

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

.

2

Convert photon energy from electron-volts to Joules

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

Since Planck's constant is given in SI units (J s), energy must be converted to Joules using

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

.

3

Calculate photon frequency using Planck's relation

f = 8.0 \times 10^{14}\text{ Hz}

Applying

f = \frac{\Delta E}{h} = \frac{5.28 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}}

yields

8.0 \times 10^{14}\text{ Hz}

.

Anahtar Kavram

Photon Emission and Energy Level Transitions

Soru 853Soru →

What is the exact numerical value of the trigonometric expression

\frac{12 \sin 30^\circ \cos 30^\circ \tan 60^\circ}{\sin^2 45^\circ + \cos^2 45^\circ + \tan^2 60^\circ}

?

Cevabı ve açıklamayı göster

Cevap: 2.25

Cevap

The exact numerical value of the expression is 2.25.

Substituting the exact values

\sin 30^\circ = \frac{1}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

, and

\tan 60^\circ = \sqrt{3}

into the numerator gives

12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

. Using the Pythagorean identity

\sin^2 45^\circ + \cos^2 45^\circ = 1

and

\tan^2 60^\circ = 3

, the denominator evaluates to

1 + 3 = 4

. Dividing 9 by 4 yields the exact value of 2.25.

Adım Adım Çözüm

1

Evaluate the trigonometric ratios for special angles and apply trigonometric identities.

\sin 30^\circ = \frac{1}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

,

\tan 60^\circ = \sqrt{3}

, and

\sin^2 45^\circ + \cos^2 45^\circ = 1

.

Standard special angle values and the fundamental Pythagorean identity

\sin^2 x + \cos^2 x = 1

simplify the expression.

2

Substitute these values into the numerator of the expression.

Numerator

= 12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

.

Multiplying the terms:

\sqrt{3} \times \sqrt{3} = 3

, and

12 \times \frac{1}{4} \times 3 = 9

.

3

Substitute these values into the denominator of the expression.

Denominator

= 1 + (\sqrt{3})^2 = 1 + 3 = 4

.

The sum

\sin^2 45^\circ + \cos^2 45^\circ = 1

combined with

\tan^2 60^\circ = 3

equals 4.

4

Divide the calculated numerator by the denominator.

\frac{9}{4} = 2.25

.

Dividing 9 by 4 gives the final decimal value 2.25.

Anahtar Kavram

Evaluation of trigonometric expressions using special angles (30°, 45°, 60°) and fundamental identities

Soru 854Soru →

An investor deposited $\text{₦}60,000$ into a financial scheme for $2$ years at an annual interest rate of $r\%$ . If the difference between the compound interest (compounded annually) and the simple interest earned over the $2$ -year period is $\text{₦}384$ , calculate the value of $r$ .

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The interest rate r is 8%

For a two-year investment period, the difference between compound interest (compounded annually) and simple interest equals the interest earned in the second year on the first year's interest, which is

P \left(\frac{r}{100}\right)^2

. Setting

60,000 \times \frac{r^2}{10,000} = 384

yields

6r^2 = 384

, giving

r^2 = 64

and

r = 8\%

.

Adım Adım Çözüm

1

Express the simple interest for 2 years in terms of r

I_{\text{SI}} = 1,200r

Simple interest is calculated directly on the principal amount for the full term.

2

Express the compound interest for 2 years in terms of r

I_{\text{CI}} = 1,200r + 6r^2

Compound interest includes interest earned on the first year's interest.

3

Set up the equation for the difference between compound and simple interest

(1,200r + 6r^2) - 1,200r = 384 \implies 6r^2 = 384

The difference between CI and SI over 2 years isolates the interest-on-interest component.

4

Solve for r

r = 8

Dividing 384 by 6 gives 64, and taking the principal square root yields 8.

Anahtar Kavram

Difference between Compound Interest and Simple Interest for 2 years

Soru 855Soru →

A liquid has a real cubic expansivity of $5.0 \times 10^{-4} \text{ K}^{-1}$ . When this liquid is heated inside a metallic vessel, its apparent cubic expansivity is determined to be $4.1 \times 10^{-4} \text{ K}^{-1}$ . What is the linear expansivity of the metallic vessel in $10^{-5} \text{ K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The linear expansivity of the metallic vessel is

3.0 \times 10^{-5} \text{ K}^{-1}

.

Real cubic expansivity of a liquid accounts for both the expansion of the liquid relative to the container and the expansion of the container itself:

\gamma_r = \gamma_a + \gamma_v

. Subtracting the apparent cubic expansivity (

4.1 \times 10^{-4} \text{ K}^{-1}

) from the real cubic expansivity (

5.0 \times 10^{-4} \text{ K}^{-1}

) gives the vessel's cubic expansivity of

0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

. Dividing this value by 3 gives the vessel's linear expansivity:

\alpha = 3.0 \times 10^{-5} \text{ K}^{-1}

.

Adım Adım Çözüm

1

Find the cubic expansivity of the vessel (

\gamma_v

)

\gamma_v = \gamma_r - \gamma_a = 5.0 \times 10^{-4} - 4.1 \times 10^{-4} = 0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the containing vessel (

\gamma_r = \gamma_a + \gamma_v

).

2

Calculate the linear expansivity of the vessel (

\alpha

)

\alpha = \frac{\gamma_v}{3} = \frac{9.0 \times 10^{-5}}{3} = 3.0 \times 10^{-5} \text{ K}^{-1}

For isotropic solids, volume (cubic) expansivity is three times the linear expansivity (

\gamma_v = 3\alpha

).

Anahtar Kavram

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

Soru 856Soru →

The cumulative frequency distribution table below shows the mass of cocoa beans (in kg) harvested by $40$ smallholder farmers in a agricultural cooperative:

Mass (kg)	Cumulative Frequency
$\leq 20$	$4$
$\leq 30$	$12$
$\leq 40$	$28$
$\leq 50$	$36$
$\leq 60$	$40$

Using linear interpolation from the cumulative frequency table, what is the median mass (in kg) of cocoa beans harvested by the farmers?

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

The median mass of cocoa beans harvested by the farmers is

35.0\text{ kg}

.

The total number of farmers is

N = 40

. The median corresponds to the

\frac{40}{2} = 20^{\text{th}}

position. From the cumulative frequency table, the

20^{\text{th}}

item lies within the

30 - 40\text{ kg}

class interval. Using the grouped median formula

\text{Median} = L + \left(\frac{\frac{N}{2} - c.f.}{f}\right) \times w

, where

L = 30

,

c.f. = 12

,

f = 28 - 12 = 16

, and

w = 10

, we obtain

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 35.0\text{ kg}

.

Adım Adım Çözüm

1

Find the median position in the cumulative frequency distribution

Median position

= \frac{N}{2} = \frac{40}{2} = 20^{\text{th}}

item

The median corresponds to the 50th percentile, which is half of the total cumulative frequency

N = 40

.

2

Locate the median class interval and extract its statistical parameters

Median class interval is

30 - 40\text{ kg}

, with lower limit

L = 30\text{ kg}

, preceding cumulative frequency

c.f. = 12

, class frequency

f = 28 - 12 = 16

, and width

w = 10\text{ kg}

The cumulative frequency just below

20

is

12

(at upper boundary

30

), and at upper boundary

40

it rises to

28

.

3

Substitute values into the grouped data median formula

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 30 + 5 = 35.0\text{ kg}

Linear interpolation estimates the exact position of the median within the median class interval.

Anahtar Kavram

Calculation of Median from Cumulative Frequency Data

Soru 857Soru →

A body is projected from ground level with an initial speed of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the time taken, in seconds, for the body to reach its maximum height. (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The time taken to reach maximum height is

2\text{ s}

.

The initial vertical velocity component is

u_y = u \sin(30^\circ) = 40 \times 0.5 = 20\text{ m/s}

. Under gravitational deceleration (

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

), the vertical speed drops to zero at maximum height after

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

.

Adım Adım Çözüm

1

Find the vertical component of initial velocity (

u_y

)

u_y = 40 \sin(30^\circ) = 20\text{ m/s}

Only the vertical component of initial velocity determines the time to reach maximum height.

2

Calculate the time to maximum height (

t

)

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

At maximum height, vertical velocity

v_y = 0

, giving

t = \frac{u_y}{g}

.

Anahtar Kavram

Time to reach maximum height in projectile motion

Soru 858Soru →

A straight line $L_1$ has the equation $3x - 4y + 5 = 0$ . A second line $L_2$ is parallel to $L_1$ and passes through the point $(6, 1)$ . What is the perpendicular distance between lines $L_1$ and $L_2$ ?

Cevabı ve açıklamayı göster

Cevap: 3.8

Cevap

The perpendicular distance between lines

L_1

and

L_2

is

3.8

units.

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Line

L_2

is parallel to

3x - 4y + 5 = 0

, so its equation is

3x - 4y + C = 0

. Substituting

(6, 1)

gives

3(6) - 4(1) + C = 0

, leading to

C = -14

. Substituting

C_1 = 5

and

C_2 = -14

into the distance formula gives

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5} = 3.8

.

Adım Adım Çözüm

1

Determine the equation of line

L_2

The equation of

L_2

is

3x - 4y - 14 = 0

Lines parallel to

3x - 4y + 5 = 0

have the form

3x - 4y + C = 0

. Substituting the point

(6, 1)

gives

3(6) - 4(1) + C = 0 \implies C = -14

.

2

Apply the parallel line distance formula

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5}

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

3

Convert fraction to decimal form

3.8

Dividing

19

by

5

yields

3.8

.

Anahtar Kavram

Perpendicular Distance Between Parallel Lines

Soru 859Soru →

A convex polygon with $n$ sides has a total interior angle sum of $1440^\circ$ . If $(n - 4)$ of its interior angles each measure $150^\circ$ , and the remaining four interior angles are equal in measure, what is the measure of one of the remaining interior angles in degrees?

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Cevap: 135

Cevap

The measure of one of the remaining interior angles is

135^\circ

.

Using the interior angle sum formula

(n - 2) \times 180^\circ = 1440^\circ

, we find

n - 2 = 8

, which means the polygon has

n = 10

sides. The number of angles measuring

150^\circ

is

10 - 4 = 6

. Their total measure is

6 \times 150^\circ = 900^\circ

. The sum of the remaining four equal angles is

1440^\circ - 900^\circ = 540^\circ

. Dividing

540^\circ

by 4 gives

135^\circ

for each remaining interior angle.

Adım Adım Çözüm

1

Calculate the total number of sides

n

of the convex polygon.

n = 10

The sum of interior angles of an

n

-sided polygon is given by

(n - 2) \times 180^\circ

. Setting

(n - 2) \times 180^\circ = 1440^\circ

gives

n - 2 = 8

, so

n = 10

.

2

Determine the number of interior angles that measure

150^\circ

each and find their combined sum.

6 angles totaling

900^\circ

There are

(n - 4) = 10 - 4 = 6

interior angles of

150^\circ

each. Their sum is

6 \times 150^\circ = 900^\circ

.

3

Calculate the sum of the remaining four equal interior angles.

540^\circ

Subtracting the sum of the known angles from the total interior angle sum yields

1440^\circ - 900^\circ = 540^\circ

.

4

Find the measure of one of the remaining four equal angles.

135^\circ

Dividing the remaining sum equally among the 4 angles gives

540^\circ / 4 = 135^\circ

.

Anahtar Kavram

Polygon interior angle sum theorem

Tahmini Süre:1m 30s

Soru 860Soru →

A metallic rod of initial length $2.0\text{ m}$ experiences a temperature increase of $50\text{ K}$ . If the linear expansivity of the metal is $1.5 \times 10^{-5}\text{ K}^{-1}$ , what is the expansion in length of the rod in millimetres?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The expansion in length of the rod is 1.5 mm.

The expansion in length is calculated using

\Delta L = L_0 \alpha \Delta T

. Substituting the values

L_0 = 2.0\text{ m}

,

\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}

, and

\Delta T = 50\text{ K}

gives

\Delta L = 1.5 \times 10^{-3}\text{ m}

, which equals

1.5\text{ mm}

.

Adım Adım Çözüm

1

Identify known variables from the problem statement.

L_0 = 2.0\text{ m}

,

\Delta T = 50\text{ K}

, and

\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}

.

Listing given physical quantities clarifies which thermal expansion formula to apply.

2

Calculate the change in length in metres using

\Delta L = L_0 \alpha \Delta T

.

\Delta L = 2.0\text{ m} \times (1.5 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 0.0015\text{ m}

.

Thermal expansion in one dimension is directly proportional to initial length, linear expansivity, and temperature change.

3

Convert the calculated expansion from metres to millimetres.

0.0015\text{ m} \times 1000\text{ mm/m} = 1.5\text{ mm}

.

The question explicitly requests the value in millimetres.

Anahtar Kavram

Linear thermal expansivity defines the fractional change in length per degree temperature change.

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