Tüm alıştırma soruları

Soru 1461Soru →

A consumer obtains a total utility ( $TU$ ) of $18$ utils, $32$ utils, $42$ utils, $48$ utils, and $50$ utils from consuming $1$ , $2$ , $3$ , $4$ , and $5$ cups of coffee, respectively. What is the marginal utility ( $MU$ ) derived from consuming the $4^{\text{th}}$ cup of coffee?

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Cevap: 6

Cevap

The marginal utility derived from consuming the 4th cup of coffee is 6 utils.

Marginal Utility (

MU

) represents the extra utility gained from consuming an additional unit of a commodity. It is computed as

MU = TU_n - TU_{n-1}

. For the 4th cup of coffee,

MU_4 = TU_4 - TU_3 = 48 - 42 = 6

utils.

Adım Adım Çözüm

1

Identify the Total Utility for the target unit and the preceding unit

TU_4 = 48

utils and

TU_3 = 42

utils

Marginal Utility measures the addition to total utility from consuming one more unit.

2

Apply the Marginal Utility formula

MU = \frac{\Delta TU}{\Delta Q}

MU_4 = \frac{48 - 42}{4 - 3} = \frac{6}{1} = 6

utils

The change in quantity is 1 unit, so

MU

is simply the difference between

TU_4

and

TU_3

.

Anahtar Kavram

Calculation of Marginal Utility from Total Utility schedule

Tahmini Süre:1m 0s

Soru 1462Soru →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a period of oscillation of $0.20\pi\text{ s}$ . What is the maximum kinetic energy of the particle in joules?

Cevabı ve açıklamayı göster

Cevap: 0.025

Cevap

The maximum kinetic energy of the particle is

0.025\text{ J}

.

The maximum kinetic energy occurs at the equilibrium position where speed reaches its maximum value

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

. Substituting

\omega = \frac{2\pi}{0.20\pi} = 10\text{ rad/s}

into the maximum speed formula gives

v_{\text{max}} = 10 \times 0.05 = 0.50\text{ m/s}

. Computing kinetic energy yields

E_k = \frac{1}{2} (0.20) (0.50)^2 = 0.025\text{ J}

.

Adım Adım Çözüm

1

Calculate angular frequency (\omega)

\omega = 10\text{ rad/s}

Using the relation

\omega = \frac{2\pi}{T}

with

T = 0.20\pi\text{ s}

.

2

Calculate maximum speed (v_{max})

v_{max} = 0.50\text{ m/s}

Maximum velocity occurs at the equilibrium position and is given by

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

.

3

Calculate maximum kinetic energy (E_k)

E_k = 0.025\text{ J}

Using the kinetic energy formula

E_k = \frac{1}{2} m v^2

at maximum velocity.

Anahtar Kavram

Energy conservation and maximum speed in Simple Harmonic Motion

Soru 1463Soru →

A poultry farm operates in a perfectly competitive market where the market price per crate of eggs is $\text{N}70$ . The farm's short-run total cost function is $TC = 2Q^2 + 10Q + 200$ , where $Q$ is the quantity of crates produced and $TC$ is the total cost in Naira. What is the maximum economic profit, in Naira, realized by the farm?

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Cevap: 250

Cevap

The maximum economic profit realized by the farm is 250 Naira.

In a perfectly competitive market, the firm is a price taker so marginal revenue equals market price (

MR = P = 70

). Equating marginal revenue to marginal cost (

MC = 4Q + 10

) gives an equilibrium output of

Q = 15

units. Substituting

Q = 15

into Total Revenue (

TR = 70 \times 15 = 1050

) and Total Cost (

TC = 2(15)^2 + 10(15) + 200 = 800

) yields an economic profit of

1050 - 800 = \text{N}250

.

Adım Adım Çözüm

1

Derive the Marginal Cost equation from Total Cost

MC = 4Q + 10

Marginal Cost represents the rate of change of Total Cost with respect to output.

2

Equate Marginal Revenue (which equals price in perfect competition) to Marginal Cost

70 = 4Q + 10 \implies Q = 15

units

A price-taking firm maximizes profit where price equals marginal cost (

P = MC

).

3

Calculate Total Revenue at optimal output level

TR = 70 \times 15 = \text{N}1050

Total revenue is the product of price per unit and total output sold.

4

Calculate Total Cost at optimal output level

TC = 2(15)^2 + 10(15) + 200 = \text{N}800

Substitute optimal output into the total cost function.

5

Subtract Total Cost from Total Revenue to find economic profit

\text{Profit} = 1050 - 800 = \text{N}250

Economic profit is the surplus remaining after deducting total explicit and fixed cost from total revenue.

Anahtar Kavram

Profit Maximization in Perfect Competition

Tahmini Süre:1m 30s

Soru 1464Soru →

A spherical black body radiator with an initial radius of $0.10\text{ m}$ emits thermal radiation at a rate of $E_1$ when its surface temperature is $400\text{ K}$ . If the radius of the sphere is doubled to $0.20\text{ m}$ and its absolute temperature is reduced to $200\text{ K}$ , what is the value of the ratio of the new rate of heat radiation to the initial rate, $\frac{E_2}{E_1}$ ?

Cevabı ve açıklamayı göster

Cevap: 0.25

Cevap

The ratio of the new rate of heat radiation to the initial rate is

0.25

.

According to the Stefan-Boltzmann law, the rate of thermal radiation emitted by a black body is directly proportional to its surface area (

A \propto r^2

) and the fourth power of its absolute temperature (

T^4

). Doubling the radius increases the surface area by a factor of

2^2 = 4

, while halving the absolute temperature decreases the radiation rate per unit area by a factor of

(1/2)^4 = 1/16

. The net ratio of the new emission rate to the initial emission rate is

4 \times (1/16) = 0.25

.

Adım Adım Çözüm

1

Apply Stefan-Boltzmann's law of radiation to formulate the rate of heat emission.

The total power radiated by a sphere of radius

r

at thermodynamic temperature

T

is given by

P = \sigma A T^4 = 4\pi \sigma r^2 T^4

, where

\sigma

is the Stefan-Boltzmann constant.

Thermal radiation emission rate depends directly on surface area (

A = 4\pi r^2

) and the fourth power of absolute temperature (

T^4

).

2

Formulate the ratio

\frac{E_2}{E_1}

using the scaled physical parameters.

\frac{E_2}{E_1} = \frac{4\pi \sigma r_2^2 T_2^4}{4\pi \sigma r_1^2 T_1^4} = \left(\frac{r_2}{r_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4

Constants

4\pi\sigma

cancel out when evaluating relative changes.

3

Substitute the given numerical ratios into the equation and compute the result.

Given

\frac{r_2}{r_1} = \frac{0.20}{0.10} = 2

and

\frac{T_2}{T_1} = \frac{200}{400} = 0.5

, we get

\frac{E_2}{E_1} = (2)^2 \times (0.5)^4 = 4 \times \frac{1}{16} = 0.25

.

Evaluating

2^2 = 4

and

(0.5)^4 = 1/16

yields a final ratio of

4/16 = 0.25

.

Anahtar Kavram

Stefan-Boltzmann Law of Radiation

Soru 1465Soru →

A light ray strikes a plane mirror. Keeping the direction of the incident ray fixed, the mirror is rotated through an angle of $15^\circ$ about an axis lying in its plane. What is the angle of rotation, in degrees, of the reflected ray?

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Cevap: 30

Cevap

The reflected ray turns through an angle of

30^\circ

.

When a plane mirror is rotated through an angle

\theta

while keeping the incident ray direction constant, the normal turns by

\theta

. This changes the angle of incidence by

\theta

and the angle of reflection by

\theta

, causing the reflected ray to rotate by a total angle of

2\theta

. For a mirror rotation of

15^\circ

, the reflected ray rotates through

2 \times 15^\circ = 30^\circ

.

Adım Adım Çözüm

1

Analyze the effect of mirror rotation on the normal line

When the plane mirror rotates by

15^\circ

, the normal to the mirror surface also rotates by

15^\circ

.

The normal line is always perpendicular to the surface of the mirror.

2

Determine the change in the angle of incidence and reflection

The angle of incidence changes by

15^\circ

, so the angle of reflection relative to the new normal also changes by

15^\circ

.

According to the law of reflection, the angle of incidence equals the angle of reflection (

i = r

).

3

Calculate the total angular deviation of the reflected ray

The total shift of the reflected ray relative to its original path is

15^\circ + 15^\circ = 30^\circ

.

The rotation of the reflected ray is twice the angle of rotation of the mirror.

Anahtar Kavram

Rotation of Reflected Ray by a Plane Mirror

Soru 1466Soru →

A glass prism with a refracting angle of $60^\circ$ has a refractive index of $\sqrt{2}$ . What is the angle of minimum deviation, in degrees, experienced by a light ray passing through this prism?

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Cevap: 30

Cevap

The angle of minimum deviation is

30^\circ

.

Using the prism minimum deviation relation

n = \frac{\sin\left((A + D_m)/2\right)}{\sin(A/2)}

, substituting

n = \sqrt{2}

and

A = 60^\circ

yields

\sin\left((60^\circ + D_m)/2\right) = \frac{1}{\sqrt{2}}

. This gives

(60^\circ + D_m)/2 = 45^\circ

, so

D_m = 30^\circ

.

Adım Adım Çözüm

1

Apply the prism minimum deviation equation.

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

This relates refractive index, prism refracting angle, and minimum deviation angle.

2

Substitute the given values

A = 60^\circ

and

n = \sqrt{2}

.

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Dividing the refracting angle

60^\circ

by 2 gives

30^\circ

for the denominator angle.

3

Calculate the numerator sine term.

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{1}{\sqrt{2}}

Multiplying

\sqrt{2}

by

\sin(30^\circ) = 0.5

gives

\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

4

Solve for the angle

D_m

.

D_m = 30^\circ

Since

\arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^\circ

, we have

\frac{60^\circ + D_m}{2} = 45^\circ

, leading to

60^\circ + D_m = 90^\circ

.

Anahtar Kavram

Minimum Deviation in Triangular Prisms

Soru 1467Soru →

A block of transparent polymer with a refractive index of $1.25$ is placed over a small mark on a table. When viewed vertically from directly above, the mark appears to be shifted upward by $3.0\text{ cm}$ . What is the real thickness of the polymer block in centimeters?

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Cevap: 15

Cevap

The real thickness of the polymer block is

15.0\text{ cm}

.

Refraction at the interface produces an upward shift

s = d\left(1 - \frac{1}{n}\right)

. Substituting

s = 3.0\text{ cm}

and

n = 1.25

gives

3.0 = d(1 - 0.80) = 0.20d

, which solves to a real thickness

d = 15.0\text{ cm}

.

Adım Adım Çözüm

1

Relate apparent depth

d'

to real depth

d

using the refractive index.

d' = \frac{d}{n}

Light refracting at the surface of a denser medium causes the object to appear at a shallower depth.

2

Express the apparent displacement (upward shift)

s

in terms of

d

and

n

.

s = d - d' = d\left(1 - \frac{1}{n}\right)

The upward shift is the vertical distance between the true position and the apparent position.

3

Substitute

s = 3.0\text{ cm}

and

n = 1.25

into the equation and solve for

d

.

3.0 = d\left(1 - \frac{1}{1.25}\right) = d(1 - 0.80) = 0.20 d \implies d = 15.0\text{ cm}

Algebraic evaluation yields the exact real thickness of the polymer block.

Anahtar Kavram

Apparent depth and upward displacement due to refraction

Soru 1468Soru →

A crate of mass $5\text{ kg}$ is pulled from rest along a smooth inclined plane tilted at $30^\circ$ to the horizontal by a constant force of $40\text{ N}$ acting parallel to the slope. What is the kinetic energy of the crate in Joules after moving a distance of $6\text{ m}$ along the incline? (Take $g = 10\text{ m s}^{-2}$ )

Cevabı ve açıklamayı göster

Cevap: 90

Cevap

The final kinetic energy of the crate is

90\text{ J}

.

The kinetic energy gained equals the net work done on the crate. The total work put in by the pulling force is

40 \times 6 = 240\text{ J}

, while the gravitational potential energy gained is

5 \times 10 \times (6 \sin 30^\circ) = 150\text{ J}

. Subtracting the potential energy gained from the total work gives a net kinetic energy of

240 - 150 = 90\text{ J}

.

Adım Adım Çözüm

1

Calculate the total work done by the applied force parallel to the incline

W_{\text{applied}} = 40\text{ N} \times 6\text{ m} = 240\text{ J}

Work done by a force is equal to force multiplied by displacement in the direction of the force.

2

Calculate the work done against gravity (potential energy gained)

\Delta PE = m g d \sin(30^\circ) = 5 \times 10 \times 6 \times 0.5 = 150\text{ J}

The height gained along an incline of length

d

and angle

\theta

is

h = d \sin\theta

.

3

Determine the net work done to find the final kinetic energy

KE = W_{\text{applied}} - \Delta PE = 240\text{ J} - 150\text{ J} = 90\text{ J}

According to the work-energy theorem, the net work done on an object equals its change in kinetic energy.

Anahtar Kavram

Work-Energy Theorem on an Inclined Plane

Soru 1469Soru →

A body of mass $0.40\text{ kg}$ is attached to a light helical spring and set into simple harmonic motion. If the system oscillates with a period of $0.40\pi\text{ s}$ , what is the force constant of the spring in $\text{N/m}$ ?

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Cevap: 10

Cevap

The force constant of the spring is

10\text{ N/m}

.

The period of a mass-spring system undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{m}{k}}

. Substituting

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

yields

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

. Dividing both sides by

2\pi

gives

0.20 = \sqrt{\frac{0.40}{k}}

. Squaring both sides yields

0.04 = \frac{0.40}{k}

, which gives

k = 10\text{ N/m}

.

Adım Adım Çözüm

1

Identify the period relationship for a spring-mass SHM system

T = 2\pi \sqrt{\frac{m}{k}}

This formula relates the period of oscillation

T

to the mass

m

and spring constant

k

.

2

Substitute the given mass and period into the equation

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

Given that

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

.

3

Simplify the equation by isolating the square root term

\sqrt{\frac{0.40}{k}} = 0.20

Dividing both sides of the equation by

2\pi

isolates the radical.

4

Square both sides and solve for the spring constant

k

k = 10\text{ N/m}

Squaring yields

0.04 = \frac{0.40}{k}

, which rearranges to

k = \frac{0.40}{0.04} = 10\text{ N/m}

.

Anahtar Kavram

Period of oscillation of a mass-spring system in Simple Harmonic Motion

Tahmini Süre:1m 30s

Soru 1470Soru →

At a certain research station, the vertical component of the Earth's magnetic field is measured as $2.4 \times 10^{-5}\text{ T}$ . If the angle of dip at this station is $37^\circ$ (given $\sin 37^\circ = 0.60$ and $\cos 37^\circ = 0.80$ ), calculate the horizontal component of the Earth's magnetic field in $\mu\text{T}$ .

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

The horizontal component of the Earth's magnetic field is

32\ \mu\text{T}

.

The horizontal component

B_h

and vertical component

B_v

of the Earth's magnetic field are related by

\tan \theta = \frac{B_v}{B_h}

, where

\theta

is the angle of dip. Given

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

and

\tan 37^\circ = 0.75

, rearranging gives

B_h = \frac{24}{0.75} = 32\ \mu\text{T}

.

Adım Adım Çözüm

1

Express the relationship between the vertical component (

B_v

), horizontal component (

B_h

), and angle of dip (

\theta

).

\tan \theta = \frac{B_v}{B_h}

By definition of the angle of dip in the magnetic meridian, the tangent of the dip angle equals the ratio of the vertical component to the horizontal component.

2

Calculate

\tan 37^\circ

using the provided trigonometric values.

\tan 37^\circ = \frac{0.60}{0.80} = 0.75

Tangent of an angle is the ratio of sine to cosine of that angle.

3

Convert

B_v

from teslas to microteslas.

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

Since

1\ \mu\text{T} = 10^{-6}\text{ T}

,

2.4 \times 10^{-5}\text{ T} = 24 \times 10^{-6}\text{ T} = 24\ \mu\text{T}

.

4

Rearrange the equation to solve for

B_h

and substitute the values.

B_h = \frac{B_v}{\tan \theta} = \frac{24\ \mu\text{T}}{0.75} = 32\ \mu\text{T}

Dividing

24

by

0.75

yields

32

.

Anahtar Kavram

Resolution of Earth's magnetic field into horizontal and vertical components

Soru 1471Soru →

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ connected in series across an AC source of RMS voltage $150\ \text{V}$ . What is the average power dissipated in the circuit in watts?

Cevabı ve açıklamayı göster

Cevap: 270

Cevap

The average power dissipated in the circuit is

270\ \text{W}

.

The total impedance of a series RLC circuit is

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega

. The RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150}{50} = 3\ \text{A}

. Because pure inductors and capacitors consume zero average power over a complete cycle, power is dissipated only across the resistor, giving

P = I_{\text{rms}}^2 R = 3^2 \times 30 = 270\ \text{W}

.

Adım Adım Çözüm

1

Calculate net reactance

X = 40\ \Omega

Inductive and capacitive reactances oppose each other in phase, so net reactance is

X_L - X_C

.

2

Calculate total circuit impedance

Z = 50\ \Omega

Resistance and net reactance add in quadrature:

Z = \sqrt{R^2 + (X_L - X_C)^2}

.

3

Calculate RMS current

I_{\text{rms}} = 3\ \text{A}

Ohm's law for AC circuits gives

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}

.

4

Calculate average power dissipated

P = 270\ \text{W}

Power is dissipated exclusively by resistance in an AC circuit:

P = I_{\text{rms}}^2 R

.

Anahtar Kavram

Power Dissipation in AC Circuits

Tahmini Süre:1m 30s

Soru 1472Soru →

A container holds a layer of water of depth $16.0\text{ cm}$ . An immiscible layer of oil of refractive index $1.20$ and thickness $6.0\text{ cm}$ floats on top of the water. If the refractive index of water is $1.33$ (or $\frac{4}{3}$ ), what is the total apparent depth, in centimeters, of a small object resting at the bottom of the container when viewed normally from directly above?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

The total apparent depth of the object when viewed normally from directly above is

17.0\text{ cm}

.

When an object at the bottom of a container is viewed normally through multiple transparent media, the overall apparent depth is the sum of the apparent depths produced by each medium individually (

d_{\text{app}} = \sum \frac{d_i}{n_i}

). Substituting the given values gives

\frac{16.0}{4/3} + \frac{6.0}{1.20} = 12.0\text{ cm} + 5.0\text{ cm} = 17.0\text{ cm}

.

Adım Adım Çözüm

1

Calculate the apparent depth of the water layer.

Apparent depth of water =

12.0\text{ cm}

Apparent depth in a single medium is given by the real depth divided by its refractive index (

d_{\text{app}} = d / n

). For water,

\frac{16.0}{4/3} = 12.0\text{ cm}

.

2

Calculate the apparent depth of the oil layer.

Apparent depth of oil =

5.0\text{ cm}

Using

d_{\text{app}} = d / n

for the oil layer,

\frac{6.0}{1.20} = 5.0\text{ cm}

.

3

Sum the apparent depths of both media.

Total apparent depth =

17.0\text{ cm}

For multiple parallel transparent layers, the total apparent depth is the sum of the apparent depths of each individual layer.

Anahtar Kavram

Apparent depth in composite media layers

Soru 1473Soru →

The saturated vapour pressure of water in an enclosed space is $25\text{ mmHg}$ . If the relative humidity within the space is measured to be $64\%$ , what is the partial pressure of the water vapour in $\text{mmHg}$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The partial pressure of the water vapour in the enclosed space is

16\text{ mmHg}

.

Relative humidity is defined as the ratio of the actual partial pressure of water vapour present in a given volume of air to the saturated vapour pressure at the same temperature, expressed as a percentage:

\text{R.H.} = \frac{P}{\text{S.V.P.}} \times 100\%

. Rearranging this equation gives

P = \frac{\text{R.H.} \times \text{S.V.P.}}{100} = \frac{64 \times 25}{100} = 16\text{ mmHg}

.

Adım Adım Çözüm

1

Identify the given parameters and state the relative humidity formula.

Relative Humidity (R.H.) =

64\%

, Saturated Vapour Pressure (S.V.P.) =

25\text{ mmHg}

.

\text{R.H.} = \frac{P}{\text{S.V.P.}} \times 100\%

.

Relative humidity relates the actual partial vapour pressure present to the maximum saturated vapour pressure possible at that temperature.

2

Substitute the values into the equation and solve for partial pressure

P

.

64 = \frac{P}{25} \times 100 \implies 64 = 4P \implies P = 16\text{ mmHg}

.

Dividing

100

by

25

yields a factor of

4

, allowing clean mental computation.

Anahtar Kavram

Relationship between relative humidity, partial vapour pressure, and saturated vapour pressure.

Soru 1474Soru →

An electric heater rated at $200\text{ W}$ is used to heat a liquid of mass $1.5\text{ kg}$ for $7\text{ minutes}$ . If the temperature of the liquid rises from $25^\circ\text{C}$ to $65^\circ\text{C}$ , assuming no heat energy is lost to the surroundings, what is the specific heat capacity of the liquid in $\text{J}\text{kg}^{-1}\text{K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 1400

Cevap

The specific heat capacity of the liquid is

1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

The electrical energy supplied over

7\text{ minutes}

(

420\text{ s}

) at a power rating of

200\text{ W}

equals

84,000\text{ J}

. Equating this total energy to thermal absorption

Q = m c \Delta T

for a

1.5\text{ kg}

mass experiencing a

40\text{ K}

temperature rise gives

c = \frac{84,000}{1.5 \times 40} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

Adım Adım Çözüm

1

Convert heating time to seconds and compute total heat energy supplied.

t = 7 \times 60 = 420\text{ s}

, so

Q = P \times t = 200 \times 420 = 84,000\text{ J}

.

Electrical power is the rate of energy transfer (

P = Q/t

), so heat energy equals power multiplied by time in seconds.

2

Calculate temperature difference.

\Delta T = 65^\circ\text{C} - 25^\circ\text{C} = 40\text{ K}

.

The temperature rise drives heat absorption according to the thermal equation.

3

Rearrange the heat energy formula

Q = m c \Delta T

to find the specific heat capacity

c

.

c = \frac{84,000}{1.5 \times 40} = \frac{84,000}{60} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

Specific heat capacity represents the heat energy required per unit mass per unit temperature change.

Anahtar Kavram

Specific heat capacity calculation using electrical energy input (

Q = P t = m c \Delta T

).

Tahmini Süre:1m 15s

Soru 1475Soru →

A particle executes simple harmonic motion with an amplitude of $0.10\text{ m}$ . At what displacement from its equilibrium position, in meters, is the kinetic energy of the particle equal to three times its potential energy?

Cevabı ve açıklamayı göster

Cevap: 0.05

Cevap

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

E_p = \frac{1}{2}kx^2

and kinetic energy is

E_k = \frac{1}{2}k(A^2 - x^2)

. Setting

E_k = 3E_p

gives

A^2 - x^2 = 3x^2

, which simplifies to

A^2 = 4x^2

, or

x = \frac{A}{2}

. Given an amplitude

A = 0.10\text{ m}

, the displacement is

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

.

Adım Adım Çözüm

1

Set up the relation between kinetic energy and potential energy using the given condition.

E_k = 3E_p \implies \frac{1}{2}k(A^2 - x^2) = 3\left(\frac{1}{2}kx^2\right)

In simple harmonic motion, energy is partitioned between kinetic and potential forms based on displacement

x

.

2

Solve the algebraic equation for displacement

x

in terms of amplitude

A

.

A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x = \frac{A}{2}

Canceling the common factor

\frac{1}{2}k

isolates the geometric parameters

A

and

x

.

3

Substitute the known amplitude value into the expression for

x

.

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

Plugging in

A = 0.10\text{ m}

yields the required displacement.

Anahtar Kavram

Energy Conservation in Simple Harmonic Motion

Soru 1476Soru →

The central bank of a country provides the following monetary statistics for a given financial period:

- Currency in circulation outside commercial banks: $\text{N}420\text{ billion}$
- Demand deposits with commercial banks: $\text{N}980\text{ billion}$
- Vault cash held by commercial banks: $\text{N}90\text{ billion}$
- Savings deposits: $\text{N}650\text{ billion}$
- Time deposits: $\text{N}350\text{ billion}$

Based on these data, what is the total broad money supply ( $M_2$ ) in this economy in billions of Naira?

Cevabı ve açıklamayı göster

Cevap: 2400

Cevap

The total broad money supply (

M_2

) in the economy is

2,400\text{ billion Naira}

.

Broad money supply (

M_2

) includes narrow money (

M_1

) plus quasi-money (savings deposits and time deposits).

M_1

is calculated as currency outside banks (

420\text{ billion Naira}

) plus demand deposits (

980\text{ billion Naira}

), totaling

1,400\text{ billion Naira}

. Adding savings deposits (

650\text{ billion Naira}

) and time deposits (

350\text{ billion Naira}

) yields

M_2 = 1,400 + 650 + 350 = 2,400\text{ billion Naira}

. Commercial bank vault cash is excluded as it is an internal reserve not available in active public circulation.

Adım Adım Çözüm

1

Calculate Narrow Money (

M_1

)

M1 = 420 + 980 = 1,400 billion Naira

M1 consists of currency in circulation outside commercial banks and demand deposits. Vault cash is excluded as it is not held by the non-bank public.

2

Calculate Quasi-Money (Near Money)

Quasi-Money = 650 + 350 = 1,000 billion Naira

Quasi-money consists of financial assets that are liquid but not directly usable as a medium of exchange, specifically savings deposits and time deposits.

3

Calculate Broad Money (

M_2

)

M2 = 1,400 + 1,000 = 2,400 billion Naira

Broad money supply combines M1 with quasi-money.

Anahtar Kavram

Broad Money Supply (M2) Calculation

Tahmini Süre:1m 30s

Soru 1477Soru →

A solid metal component of mass $0.6\text{ kg}$ absorbs $5400\text{ J}$ of thermal energy when its temperature increases from $30^\circ\text{C}$ to $75^\circ\text{C}$ . Calculate the heat capacity of the component in $\text{J K}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: 120

Cevap

The heat capacity of the metal component is

120\text{ J K}^{-1}

.

Heat capacity (

C

) is calculated by dividing the total thermal energy absorbed (

Q = 5400\text{ J}

) by the resulting temperature change (

\Delta T = 75^\circ\text{C} - 30^\circ\text{C} = 45\text{ K}

). Performing the division yields

C = \frac{5400}{45} = 120\text{ J K}^{-1}

. Note that the mass (

0.6\text{ kg}

) is not included in the denominator because heat capacity refers to the entire object, unlike specific heat capacity.

Adım Adım Çözüm

1

Determine the change in temperature

\Delta T = 75^\circ\text{C} - 30^\circ\text{C} = 45\text{ K}

Heat capacity depends on the temperature difference through which heat energy is absorbed.

2

Apply the definition of heat capacity

C = \frac{Q}{\Delta T}

Heat capacity (

C

) is defined as the thermal energy required to raise the temperature of the entire body by

1\text{ K}

(or

1^\circ\text{C}

).

3

Substitute the given thermal energy and temperature change

C = \frac{5400\text{ J}}{45\text{ K}} = 120\text{ J K}^{-1}

Dividing total energy by total temperature rise gives heat capacity.

Anahtar Kavram

Heat capacity (

C

) represents the thermal energy needed to change a body's temperature by one unit (

C = Q / \Delta T

), whereas specific heat capacity (

c

) is heat capacity per unit mass (

c = C / m

).

Soru 1478Soru →

A planet has a mass equal to $8$ times the mass of the Earth and a radius equal to $2$ times the radius of the Earth. If the acceleration due to gravity on the surface of the Earth is $10\text{ m/s}^2$ , what is the acceleration due to gravity on the surface of the planet in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The acceleration due to gravity on the surface of the planet is

20\text{ m/s}^2

.

The acceleration due to gravity at the surface of a spherical body is given by

g = \frac{GM}{R^2}

. When the mass is multiplied by

8

and the radius is multiplied by

2

, the new acceleration becomes

g_p = \frac{G(8M)}{(2R)^2} = \frac{8}{4}\frac{GM}{R^2} = 2g

. Substituting

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

gives

20\text{ m/s}^2

.

Adım Adım Çözüm

1

Write the expression for acceleration due to gravity at the surface of Earth.

g_e = \frac{GM_e}{R_e^2} = 10\text{ m/s}^2

Gravitational field strength at the surface depends directly on body mass and inversely on the square of radius.

2

Set up the ratio equation for the planet's gravitational acceleration using relative mass and radius values.

g_p = \frac{G(8M_e)}{(2R_e)^2} = \frac{8}{4} \left(\frac{GM_e}{R_e^2}\right) = 2g_e

Squaring the radius multiplier of

2

yields

4

in the denominator, while the numerator increases by a factor of

8

.

3

Calculate the final numeric value.

g_p = 2 \times 10\text{ m/s}^2 = 20\text{ m/s}^2

Multiplying Earth's value by the combined ratio of

2

gives the answer.

Anahtar Kavram

Gravitational field strength on planetary surfaces (

g = \frac{GM}{R^2}

)

Soru 1479Soru →

A radioactive parent nucleus of Uranium-238 ( $^{238}_{92}\text{U}$ ) decays through a sequence of natural radioactive emissions to become a stable daughter nucleus of Lead-206 ( $^{206}_{82}\text{Pb}$ ). How many $\beta^-$ -particles are emitted in total during this decay process?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

The decrease in mass number from

238

to

206

is

32

, which corresponds to

8

alpha emissions (

32 / 4 = 8

). Emitting

8

alpha particles reduces the atomic number from

92

to

76

(

92 - 16 = 76

). To reach the atomic number of Lead (

82

), the atomic number must be increased by

6

(

82 - 76 = 6

), which requires emitting

6

beta-minus particles.

Adım Adım Çözüm

1

Calculate the total change in mass number and deduce the number of alpha particles emitted.

The total mass number decreases by

238 - 206 = 32

. Each alpha particle (

^{4}_{2}\text{He}

) decreases the mass number by

4

, while beta particles have a mass number of

0

. Therefore, the number of alpha particles is

\frac{32}{4} = 8

.

Beta particles carry negligible mass compared to nucleons, making alpha emission the sole mechanism for mass number reduction.

2

Calculate the intermediate atomic number after all alpha decays.

Each alpha particle removes

2

units of atomic number. The total reduction in atomic number due to

8

alpha particles is

8 \times 2 = 16

. The intermediate atomic number is

92 - 16 = 76

.

Alpha particles consist of 2 protons and 2 neutrons.

3

Determine the number of beta-minus particles required to reach the final atomic number.

The final atomic number of Lead-206 is

82

. The atomic number must increase by

82 - 76 = 6

. Since each

\beta^-

-particle increases the atomic number by

1

, exactly

6

beta-minus particles are emitted.

A beta-minus decay converts a neutron into a proton, increasing the atomic number by 1 without altering the mass number.

Anahtar Kavram

Balancing mass numbers and atomic numbers in radioactive decay series

Soru 1480Soru →

Two plane mirrors are inclined to each other at an angle of $60^\circ$ . A ray of light strikes the first mirror at an angle of incidence of $40^\circ$ and subsequently undergoes reflection at the second mirror. What is the total angle of deviation, in degrees, produced in the ray after the two successive reflections?

Cevabı ve açıklamayı göster

Cevap: 240

Cevap

The total deviation produced in the light ray after successive reflections from both inclined plane mirrors is

240^\circ

.

When a light ray undergoes successive reflections at two plane mirrors inclined at an angle

\theta

, the total deviation

D

experienced by the ray is given by

D = 360^\circ - 2\theta

. Substituting

\theta = 60^\circ

yields

D = 360^\circ - 2(60^\circ) = 240^\circ

. Step-by-step ray tracing confirms

d_1 = 180^\circ - 2(40^\circ) = 100^\circ

and

d_2 = 180^\circ - 2(20^\circ) = 140^\circ

, giving a total deviation of

100^\circ + 140^\circ = 240^\circ

.

Adım Adım Çözüm

1

Calculate the deviation at the first mirror

d_1 = 100^\circ

The angle of deviation for a single reflection at a plane mirror is

d = 180^\circ - 2i

. For

i_1 = 40^\circ

,

d_1 = 180^\circ - 80^\circ = 100^\circ

.

2

Find the angle of incidence on the second mirror using geometry

i_2 = 20^\circ

The glancing angle on the first mirror is

g_1 = 90^\circ - 40^\circ = 50^\circ

. In the triangle formed by the ray path and the two mirrors inclined at

60^\circ

, the glancing angle on the second mirror is

g_2 = 180^\circ - (60^\circ + 50^\circ) = 70^\circ

. Thus, the angle of incidence on the second mirror is

i_2 = 90^\circ - 70^\circ = 20^\circ

.

3

Calculate the deviation at the second mirror and total deviation

d_2 = 140^\circ

,

D = 240^\circ

The deviation at the second mirror is

d_2 = 180^\circ - 2(20^\circ) = 140^\circ

. Since both reflections rotate the ray in the same sense, the total deviation is

D = d_1 + d_2 = 100^\circ + 140^\circ = 240^\circ

.

Anahtar Kavram

Total deviation produced by reflection from two inclined plane mirrors is independent of the initial angle of incidence and is given by

D = 360^\circ - 2\theta

, where

\theta

is the angle of inclination between the mirrors.

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