Tüm alıştırma soruları

Soru 4381Soru →

A boarding school store has sufficient provisions to feed $120$ students for $25$ days. After $5$ days of regular consumption, $30$ additional students join the school. Assuming the daily consumption rate per student remains unchanged, for how many more days will the remaining provisions last?

16\text{ days}

20\text{ days}

25\text{ days}

24\text{ days}

Cevabı ve açıklamayı göster

Cevap: $16\text{ days}$

Cevap

The remaining provisions will last for

16\text{ days}

.

After 5 days, the remaining food is sufficient to feed 120 students for 20 more days, giving a total of

120 \times 20 = 2,400

student-days of provisions. When 30 new students join, the group increases to 150 students. Dividing 2,400 student-days by 150 students gives 16 days.

Adım Adım Çözüm

1

Calculate the remaining days of food for the original number of students

\text{Remaining days} = 25 - 5 = 20\text{ days}

Food has already been consumed for 5 days out of the original 25 days.

2

Determine the total remaining food capacity in student-days

\text{Remaining supply} = 120 \text{ students} \times 20 \text{ days} = 2,400 \text{ student-days}

One student-day represents the amount of food consumed by one student in one day.

3

Calculate the new total number of students

\text{Total students} = 120 + 30 = 150 \text{ students}

30 additional students joined the original 120 students.

4

Find the number of days the remaining food will last the new total number of students

\text{Days} = \frac{2,400 \text{ student-days}}{150 \text{ students}} = 16 \text{ days}

Since consumption rate is inversely proportional to the number of consumers, divide total student-days by the total student count.

Anahtar Kavram

Inverse Proportion and Consumption Rate

Soru 4382Soru →

A particle carrying a positive charge of $3.2 \times 10^{-19}\text{ C}$ moves with a velocity of $4.0 \times 10^6\text{ m/s}$ into a uniform magnetic field of magnetic flux density $0.50\text{ T}$ . If the velocity vector of the particle makes an angle of $30^\circ$ with the direction of the magnetic field, what is the magnitude of the magnetic force exerted on the particle?

3.2 \times 10^{-13}\text{ N}

6.4 \times 10^{-13}\text{ N}

5.5 \times 10^{-13}\text{ N}

1.6 \times 10^{-13}\text{ N}

Cevabı ve açıklamayı göster

Cevap: $3.2 \times 10^{-13}\text{ N}$

Cevap

The magnitude of the magnetic force exerted on the particle is

3.2 \times 10^{-13}\text{ N}

.

The magnetic force on a charged particle moving through a magnetic field is given by

F = q v B \sin\theta

. Substituting

q = 3.2 \times 10^{-19}\text{ C}

,

v = 4.0 \times 10^6\text{ m/s}

,

B = 0.50\text{ T}

, and

\sin(30^\circ) = 0.50

yields

F = 3.2 \times 10^{-13}\text{ N}

.

Adım Adım Çözüm

1

Identify the given physical quantities

q = 3.2 \times 10^{-19}\text{ C}

,

v = 4.0 \times 10^6\text{ m/s}

,

B = 0.50\text{ T}

,

\theta = 30^\circ

Extract values needed for the magnetic force formula.

2

Apply the magnetic force formula for a moving charge

F = q v B \sin\theta

The force experienced by a moving point charge in a uniform magnetic field depends on charge, speed, field strength, and the angle between velocity and field.

3

Substitute the values and calculate

F = (3.2 \times 10^{-19}) \times (4.0 \times 10^6) \times 0.50 \times \sin(30^\circ) = (6.4 \times 10^{-13}) \times 0.50 = 3.2 \times 10^{-13}\text{ N}

Since

\sin(30^\circ) = 0.5

, evaluating the expression yields

3.2 \times 10^{-13}\text{ N}

.

Anahtar Kavram

Magnetic Force on a Moving Charge (

F = q v B \sin\theta

)

Soru 4383Soru →

A coil consisting of $50$ turns is placed in a region of changing magnetic field. If the magnetic flux passing through the coil increases uniformly from $0.2\text{ Wb}$ to $0.6\text{ Wb}$ in $2.0\text{ s}$ , what is the magnitude of the induced electromotive force in the coil in volts?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The magnitude of the induced electromotive force in the coil is

10\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force

E

is proportional to the number of turns

N

and the rate of change of magnetic flux

\frac{\Delta \Phi}{\Delta t}

. Given

N = 50

,

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

, and

\Delta t = 2.0\text{ s}

, substituting these into

E = N \frac{\Delta \Phi}{\Delta t}

gives

E = 50 \times \frac{0.4}{2.0} = 10\text{ V}

.

Adım Adım Çözüm

1

Determine the change in magnetic flux through the coil

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

Induction depends on the change in magnetic flux over time.

2

Apply Faraday's law of electromagnetic induction to solve for the induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 50 \times \frac{0.4\text{ Wb}}{2.0\text{ s}} = 10\text{ V}

Faraday's law states that the induced e.m.f. is equal to the product of the number of turns and the rate of change of flux.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction

Soru 4384Soru →

A uniform wire of length $0.60\text{ m}$ and linear mass density $4.0 \times 10^{-3}\text{ kg/m}$ is fixed at both ends under a tension of $360\text{ N}$ . When plucked, the wire vibrates in its second overtone. This frequency is found to be in resonance with the first overtone of an air column in a pipe closed at one end. Taking the speed of sound in air as $340\text{ m/s}$ , what is the length of the pipe?

0.17\text{ m}

0.34\text{ m}

0.51\text{ m}

0.68\text{ m}

Cevabı ve açıklamayı göster

Cevap: $0.34\text{ m}$

Cevap

The length of the pipe is

0.34\text{ m}

.

The correct answer of

0.34\text{ m}

is obtained by finding the wave speed on the string (

300\text{ m/s}

), computing its 3rd harmonic frequency (

750\text{ Hz}

), and equating this to the 3rd harmonic frequency formula for a closed pipe (

f = \frac{3 v_a}{4 L_p}

).

Adım Adım Çözüm

1

Calculate the speed of transverse waves on the stretched string.

v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{360\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{90000} = 300\text{ m/s}

Wave speed on a stretched string depends on tension and linear mass density.

2

Determine the fundamental frequency and the 2nd overtone frequency of the string.

Fundamental frequency

f_{1,s} = \frac{v_s}{2 L_s} = \frac{300}{2(0.60)} = 250\text{ Hz}

. Second overtone is the 3rd harmonic (

n=3

), so

f = 3 \times 250 = 750\text{ Hz}

.

For a string fixed at both ends, harmonics are integer multiples of the fundamental, and the 2nd overtone corresponds to

n=3

.

3

Relate the resonance frequency to the length of the closed pipe.

First overtone of a closed pipe is its 3rd harmonic (

m=3

):

f = \frac{3 v_a}{4 L_p} \implies 750 = \frac{3(340)}{4 L_p}

.

Pipes closed at one end only produce odd harmonics (

1, 3, 5, \dots

), so the 1st overtone is

m=3

.

4

Solve for the pipe length

L_p

.

L_p = \frac{3(340)}{4(750)} = \frac{1020}{3000} = 0.34\text{ m}

Algebraic rearrangement yields the required physical length.

Anahtar Kavram

Harmonics and overtones in vibrating strings and closed air columns under resonance

Soru 4385Soru →

Atoms of the same chemical element that possess the same atomic number but differ in their mass number are known as which of the following?

Isotopes

Allotropes

Isomers

Isobars

Cevabı ve açıklamayı göster

Cevap: Isotopes

Cevap

Atoms of the same element with identical atomic numbers but different mass numbers are called isotopes.

Isotopes are species of the same chemical element (having the same number of protons and atomic number

Z

) that contain different numbers of neutrons, resulting in different mass numbers

A

.

Adım Adım Çözüm

1

Define atomic number and mass number relations in atoms of the same element

The atomic number represents the number of protons, which determines the element's identity. The mass number is the total count of protons and neutrons.

Understanding subatomic composition is essential to identifying relationship terminology.

2

Identify the relationship where atomic number remains fixed while mass number varies

Variation in mass number among atoms of the same element occurs due to a differing number of neutrons in their nuclei.

Differing neutron counts in identical chemical elements define isotopy.

Anahtar Kavram

Subatomic Particles and Isotopy

Tahmini Süre:45s

Soru 4386Soru →

Match each vibrating system mode on the left with the correct relationship between its standing wavelength ( $\lambda$ ) and length ( $L$ ) on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The mode descriptions match their standing wavelength expressions as follows: Pipe closed at one end in its first overtone matches

\lambda = \frac{4L}{3}

; Pipe open at both ends in its first overtone matches

\lambda = L

; Stretched string in its second overtone matches

\lambda = \frac{2L}{3}

; Pipe closed at one end in its fundamental mode matches

\lambda = 4L

.

Each pair correctly links the specified boundary condition and mode of vibration to its mathematical relationship between wavelength

\lambda

and physical length

L

.

Adım Adım Çözüm

1

Identify boundary conditions and available harmonics for each vibrating system.

Closed pipes support odd harmonics only (

n = 1, 3, 5, \dots

) with

L = \frac{n\lambda}{4}

. Open pipes and fixed strings support all integer harmonics (

n = 1, 2, 3, \dots

) with

L = \frac{n\lambda}{2}

.

Boundary conditions constrain node and antinode positions, determining allowed harmonic modes.

2

Determine the specific harmonic number

n

corresponding to each specified overtone.

First overtone of closed pipe

\rightarrow n = 3

; First overtone of open pipe

\rightarrow n = 2

; Second overtone of fixed string

\rightarrow n = 3

; Fundamental of closed pipe

\rightarrow n = 1

.

Overtones are higher resonant modes above the fundamental frequency.

3

Solve for wavelength

\lambda

in terms of system length

L

for each item.

For

n = 3

(closed pipe):

L = \frac{3\lambda}{4} \Rightarrow \lambda = \frac{4L}{3}

. For

n = 2

(open pipe):

L = \lambda \Rightarrow \lambda = L

. For

n = 3

(fixed string):

L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3}

. For

n = 1

(closed pipe):

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

.

Rearranging each expression establishes the correct matching pair.

Anahtar Kavram

Boundary conditions and harmonic wavelength relations in pipes and vibrating strings

Soru 4387Soru →

A shell of total mass $5.0\text{ kg}$ is moving horizontally with a velocity of $20\text{ m s}^{-1}$ when an internal explosion splits it into two fragments. One fragment of mass $2.0\text{ kg}$ is propelled backward along the original path at a speed of $10\text{ m s}^{-1}$ . What is the magnitude of the velocity, in $\text{m s}^{-1}$ , of the second fragment immediately after the explosion?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The magnitude of the velocity of the second fragment immediately after the explosion is

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

.

According to the Law of Conservation of Linear Momentum, the total momentum of a system remains constant when no net external horizontal force acts on it. Taking the original direction of motion as positive, the initial momentum is

100\text{ kg m s}^{-1}

. Since the

2.0\text{ kg}

fragment moves backward at

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

, its momentum is

-20\text{ kg m s}^{-1}

. For total momentum to remain

+100\text{ kg m s}^{-1}

, the remaining

3.0\text{ kg}

fragment must carry a momentum of

+120\text{ kg m s}^{-1}

, which corresponds to a velocity of

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

.

Adım Adım Çözüm

1

Calculate the initial momentum of the shell prior to the explosion.

p_i = 5.0\text{ kg} \times 20\text{ m s}^{-1} = 100\text{ kg m s}^{-1}

in the initial forward direction.

Before the internal explosion, the system consists of a single mass moving with a constant velocity.

2

Apply the Law of Conservation of Linear Momentum taking vector direction into account.

p_i = m_1 v_1 + m_2 v_2 \implies 100 = 2.0(-10) + 3.0 v_2

In the absence of external forces, total momentum is conserved. The fragment propelled backward takes a negative sign relative to the initial forward motion.

3

Solve the algebraic equation for the unknown velocity

v_2

.

100 + 20 = 3.0 v_2 \implies 120 = 3.0 v_2 \implies v_2 = 40\text{ m s}^{-1}

Isolating

v_2

gives the forward velocity magnitude of the remaining

3.0\text{ kg}

piece.

Anahtar Kavram

Conservation of Linear Momentum in Explosions (1D Vector Sign Convention)

Soru 4388Soru →

An X-ray tube operates at an accelerating potential of $50.0\text{ kV}$ . Given Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ , the speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , and the elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ , what is the minimum wavelength $\lambda_{\min}$ of the emitted continuous X-rays, and how does this cutoff wavelength respond to an increase in the anode potential?

2.49 \times 10^{-11}\text{ m}

, and

\lambda_{\min}

decreases when the anode potential is increased.

2.49 \times 10^{-8}\text{ m}

, and

\lambda_{\min}

decreases when the anode potential is increased.

2.49 \times 10^{-11}\text{ m}

, and

\lambda_{\min}

remains unchanged because

\lambda_{\min}

depends only on the filament current.

4.02 \times 10^{-11}\text{ m}

, and

\lambda_{\min}

increases when the anode potential is increased.

Cevabı ve açıklamayı göster

Cevap: $2.49 \times 10^{-11}\text{ m}$ , and $\lambda_{\min}$ decreases when the anode potential is increased.

Cevap

The minimum wavelength is

2.49 \times 10^{-11}\text{ m}

, and

\lambda_{\min}

decreases when the anode potential is increased.

By the Duane-Hunt law, the maximum photon energy produced by electron impact equals the kinetic energy of the incident electron:

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

. Substituting

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

,

c = 3.00 \times 10^8\text{ m s}^{-1}

,

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

, and

V = 5.00 \times 10^4\text{ V}

yields

\lambda_{\min} = 2.49 \times 10^{-11}\text{ m}

. Because

\lambda_{\min}

is inversely proportional to

V

, increasing the anode potential decreases the minimum wavelength, making the beam more penetrating (harder).

Adım Adım Çözüm

1

Convert the operating voltage to standard SI units (volts).

V = 50.0\text{ kV} = 50.0 \times 10^3\text{ V} = 5.00 \times 10^4\text{ V}

.

Equations involving fundamental constants require input quantities in SI base units.

2

Apply the Duane-Hunt law for maximum photon energy / minimum wavelength in continuous X-ray production.

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

.

The maximum kinetic energy of an accelerating electron is completely converted into a single photon of minimum wavelength.

3

Substitute the physical values into the minimum wavelength formula.

$\lambda_{\min} = \frac{(6.63 \times 10^{-34}\text{ J s})(3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C})(5.00 \times 10^4\text{ V})} = \frac{1.989 \times 10^{-25}}{8.00 \times 10^{-15}} = 2.48625 \times 10^{-11}\text{ m} \approx 2.49 \times 10^{-11}\text{ m}$.

Direct numerical evaluation yields the minimum cutoff wavelength.

4

Analyze the functional relationship between anode potential

V

and cutoff wavelength

\lambda_{\min}

.

Since

\lambda_{\min} \propto \frac{1}{V}

, an increase in

V

results in a decrease in

\lambda_{\min}

.

Higher potential imparts greater kinetic energy to striking electrons, enabling emission of higher-frequency (shorter-wavelength) photons.

Anahtar Kavram

Duane-Hunt Law and Control Parameters of X-ray Production

Tahmini Süre:2m 0s

Soru 4389Soru →

A light ray strikes the first face of a glass prism of refracting angle $30^\circ$ at normal incidence ( $i = 0^\circ$ ). If the refractive index of the glass is $1.50$ , calculate the angle of emergence, in degrees, as the ray leaves the second face into air. (Take $\arcsin(0.75) = 48.6^\circ$ )

Cevabı ve açıklamayı göster

Cevap: 48.6

Cevap

The angle of emergence of the light ray as it exits the prism is

48.6^\circ

.

Because the ray is incident normally at the first surface, it continues undeviated into the glass (

r_1 = 0^\circ

). By prism geometry, the angle of incidence at the second face is equal to the apex angle of the prism (

r_2 = A = 30^\circ

). Applying Snell's Law at the glass-air boundary gives

1.50 \sin(30^\circ) = 1.00 \sin(e)

, leading to

\sin(e) = 0.75

, which evaluates to an emergent angle of

48.6^\circ

.

Adım Adım Çözüm

1

Determine the angle of refraction at the first surface

r_1 = 0^\circ

Light entering a surface normally (

i_1 = 0^\circ

) passes straight through without bending.

2

Find the angle of incidence at the second surface inside the prism using prism geometry

r_2 = 30^\circ

For any triangular prism, the refracting angle

A = r_1 + r_2

. Since

r_1 = 0^\circ

,

r_2 = A = 30^\circ

.

3

Apply Snell's Law at the second interface (glass to air)

\sin(e) = 0.75

n_{\text{glass}} \sin(r_2) = n_{\text{air}} \sin(e) \implies 1.50 \times \sin(30^\circ) = 1.00 \times \sin(e)

.

4

Calculate the emergent angle

e

e = 48.6^\circ

Taking the inverse sine of

0.75

yields

e = \arcsin(0.75) = 48.6^\circ

.

Anahtar Kavram

Prism Geometry and Snell's Law Refraction

Soru 4390Soru →

A converging lens has a focal length of $25\text{ cm}$ . What is the power of the lens in dioptres?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The power of the lens is

4\text{ D}

.

The power

P

of a lens in dioptres is defined as the reciprocal of its focal length

f

expressed in meters (

P = \frac{1}{f}

). Expressing

25\text{ cm}

in meters gives

0.25\text{ m}

. Substituting this value yields

P = \frac{1}{0.25\text{ m}} = 4\text{ D}

.

Adım Adım Çözüm

1

Convert the focal length from centimeters to meters

f = 25\text{ cm} = 0.25\text{ m}

The unit of dioptre (

D

) is defined as reciprocal meters (

m^{-1}

), so focal length must be in meters.

2

Apply the lens power formula

P = \frac{1}{f}

and compute the value

P = \frac{1}{0.25} = 4\text{ D}

The power of a converging lens is positive and equal to the reciprocal of its focal length in meters.

Anahtar Kavram

Power of a Lens

Soru 4391Soru →

A block is pulled along a smooth horizontal table by a horizontal force of $30\text{ N}$ through a displacement of $4\text{ m}$ . At the same time, a vertical upward force of $40\text{ N}$ acts on the block as it moves. What is the total work done on the block?

120\text{ J}

280\text{ J}

160\text{ J}

7.5\text{ J}

Cevabı ve açıklamayı göster

Cevap: $120\text{ J}$

Cevap

The total work done on the block is

120\text{ J}

.

Work done is given by

W = F d \cos\theta

. The horizontal force acts along the line of motion (

\theta = 0^\circ

), performing

30\text{ N} \times 4\text{ m} = 120\text{ J}

of work. The vertical force acts at an angle of

90^\circ

to the horizontal displacement, performing zero work because

\cos(90^\circ) = 0

. Thus, the total work done is

120\text{ J}

.

Adım Adım Çözüm

1

Identify the formula for work done by a constant force

W = F \cdot d \cdot \cos(\theta)

Work is defined as the scalar product of force and displacement vectors.

2

Calculate work done by the horizontal force

W_{\text{horizontal}} = 30\text{ N} \times 4\text{ m} \times \cos(0^\circ) = 120\text{ J}

The horizontal force is in the exact direction of motion (

\theta = 0^\circ

).

3

Calculate work done by the vertical force

W_{\text{vertical}} = 40\text{ N} \times 4\text{ m} \times \cos(90^\circ) = 0\text{ J}

The vertical force is perpendicular to the horizontal displacement (

\theta = 90^\circ

), so

\cos(90^\circ) = 0

.

4

Sum the work done by all forces

W_{\text{total}} = 120\text{ J} + 0\text{ J} = 120\text{ J}

Work is a scalar quantity, so total work is the algebraic sum of individual work values.

Anahtar Kavram

Work done by perpendicular forces is zero

Tahmini Süre:45s

Soru 4392Soru →

Match each displacement vector combination on the left with its corresponding resultant displacement magnitude or vector on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The correct pairs correspond as follows:

3\text{ m}

East and

4\text{ m}

North matches a resultant magnitude of

5\text{ m}

;

5\text{ m}

East and

12\text{ m}

South matches a resultant magnitude of

13\text{ m}

;

8\text{ m}

East and

6\text{ m}

West matches a resultant displacement of

2\text{ m}

East; and

9\text{ m}

North and

12\text{ m}

East matches a resultant magnitude of

15\text{ m}

.

Each vector combination is resolved according to its directional alignment: perpendicular displacements require the Pythagorean theorem (

R = \sqrt{A^2 + B^2}

), whereas anti-parallel collinear displacements require vector subtraction.

Adım Adım Çözüm

1

Identify orthogonal vector scenarios

Perpendicular displacement vectors form right-angled triangles.

Directions such as East-North, East-South, and North-East are at

90^\circ

relative to one another.

2

Calculate magnitudes for orthogonal pairs using the Pythagorean theorem

For

3\text{ m}

and

4\text{ m}

:

\sqrt{3^2 + 4^2} = 5\text{ m}

. For

5\text{ m}

and

12\text{ m}

:

\sqrt{5^2 + 12^2} = 13\text{ m}

. For

9\text{ m}

and

12\text{ m}

:

\sqrt{9^2 + 12^2} = 15\text{ m}

.

The magnitude of two perpendicular vectors

\vec{A}

and

\vec{B}

is given by

R = \sqrt{A^2 + B^2}

.

3

Calculate net displacement for opposite collinear vectors

For

8\text{ m}

East and

6\text{ m}

West:

8 - 6 = 2\text{ m}

East.

Vectors pointing in opposite directions along the same axis subtract algebraically, retaining the direction of the vector with the greater magnitude.

Anahtar Kavram

Addition of Perpendicular and Collinear Displacement Vectors

Tahmini Süre:1m 30s

Soru 4393Soru →

A metallic wire has a resistance of $10.0\,\Omega$ at $0^\circ\text{C}$ and a temperature coefficient of resistance $\alpha = 4.0 \times 10^{-3}\,\text{K}^{-1}$ . The wire is uniformly stretched at constant temperature until its length is doubled while maintaining constant mass and volume. It is subsequently heated to $50^\circ\text{C}$ . What is the electric current, in Amperes, that flows through the wire when a potential difference of $120.0\,\text{V}$ is applied across its ends?

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The electric current passing through the heated, stretched wire is 2.5 A.

When a wire of initial resistance 10.0 ohms is stretched to twice its original length, conservation of volume requires its cross-sectional area to halve, which quadruples its resistance to 40.0 ohms at 0 °C. Heating the wire by 50 K increases its resistance by a factor of (1 + 0.004 * 50) = 1.2, producing a final resistance of 48.0 ohms. Applying a 120.0 V potential difference across 48.0 ohms results in a current of 2.5 A.

Adım Adım Çözüm

1

Determine resistance change due to wire stretching

R_0' = 40.0 ohms at 0 °C

Uniform stretching conserves total volume (V = A * L). Doubling length halves area, making resistance increase by a factor of 2^2 = 4.

2

Calculate resistance at 50 °C using temperature coefficient

R(50 °C) = 48.0 ohms

Resistance increases linearly with temperature: R(T) = R_0'(1 + alpha * Delta T).

3

Apply Ohm's law to find current

I = 2.5 A

Current is given by potential difference divided by total resistance at the operational temperature.

Anahtar Kavram

Resistance variation with geometric stretching and temperature coefficient of resistance

Soru 4394Soru →

A radio transmitter emits electromagnetic waves with a frequency of $1.0 \times 10^8\text{ Hz}$ . Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m s}^{-1}$ , what is the wavelength of the emitted wave?

3.0\text{ m}

3.0 \times 10^{16}\text{ m}

0.33\text{ m}

3.0 \times 10^{-1}\text{ m}

Cevabı ve açıklamayı göster

Cevap: $3.0\text{ m}$

Cevap

The wavelength of the emitted radio wave is

3.0\text{ m}

.

According to the wave equation

c = f \lambda

, the wavelength is found by dividing the speed of light by frequency:

\lambda = \frac{3.0 \times 10^8\text{ m s}^{-1}}{1.0 \times 10^8\text{ Hz}} = 3.0\text{ m}

.

Adım Adım Çözüm

1

Identify the given parameters and fundamental formula

Speed of light

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

and frequency

f = 1.0 \times 10^8\text{ Hz}

. The electromagnetic wave relation is

c = f \lambda

.

All electromagnetic waves travel at the speed of light in a vacuum.

2

Rearrange the formula to solve for wavelength

\lambda = \frac{c}{f}

Dividing both sides of the wave equation by frequency isolates wavelength.

3

Substitute values into the equation and compute the result

\lambda = \frac{3.0 \times 10^8}{1.0 \times 10^8} = 3.0\text{ m}

Performing the division yields the correct wavelength.

Anahtar Kavram

Relationship between wave speed, frequency, and wavelength for electromagnetic radiation in a vacuum.

Soru 4395Soru →

A boat moves due north across a river at a speed of $4.0\text{ m s}^{-1}$ relative to the water. If the river current flows due east at a speed of $3.0\text{ m s}^{-1}$ , what is the magnitude of the resultant velocity of the boat relative to the riverbank?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $5.0\text{ m s}^{-1}$

Cevap

The magnitude of the resultant velocity of the boat relative to the riverbank is

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

.

The resultant velocity magnitude is found by applying vector addition for perpendicular components:

v_{resultant} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

Adım Adım Çözüm

1

Identify the given velocity components and their directions

Northward velocity component

v_y = 4.0\text{ m s}^{-1}

, Eastward velocity component

v_x = 3.0\text{ m s}^{-1}

.

The motion of the boat and the movement of the river current act in perpendicular directions (

90^\circ

to each other).

2

Calculate the resultant magnitude using vector addition (Pythagorean theorem)

v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

For two orthogonal vectors, the magnitude of the vector sum equals the hypotenuse of the right triangle formed by the vector components.

Anahtar Kavram

Vector Addition of Perpendicular Quantities

Tahmini Süre:1m 0s

Soru 4396Soru →

A rectangular coil of $200$ turns has dimensions $0.15\text{ m}$ by $0.10\text{ m}$ . The coil is placed with its plane perpendicular to a uniform magnetic field of $0.40\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.06\text{ s}$ , calculate the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts.

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The magnitude of the average induced electromotive force is

40\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force (e.m.f.) is given by

\mathcal{E} = N \left| \frac{\Delta \Phi}{\Delta t} \right|

. Since the coil is initially perpendicular to the magnetic field

B

, the initial flux per turn is

\Phi_i = B A

. When the field is completely reversed, the final flux becomes

\Phi_f = -B A

, giving a magnitude of flux change per turn of

|\Delta \Phi| = B A - (-B A) = 2 B A

. Substituting

N = 200

,

A = 0.015\text{ m}^2

,

B = 0.40\text{ T}

, and

\Delta t = 0.06\text{ s}

gives

\mathcal{E} = 200 \times \frac{2 \times 0.40 \times 0.015}{0.06} = 40\text{ V}

.

Adım Adım Çözüm

1

Calculate the cross-sectional area of the rectangular coil

A = 0.15\text{ m} \times 0.10\text{ m} = 0.015\text{ m}^2

The area is required to determine the magnetic flux passing through the coil.

2

Compute the initial magnetic flux per turn

\Phi_i = B A = 0.40 \times 0.015 = 0.006\text{ Wb}

Magnetic flux is defined as the product of magnetic field strength and area when perpendicular.

3

Calculate the change in flux per turn when the magnetic field reverses direction

|\Delta \Phi| = \Phi_i - (-\Phi_i) = 2 \Phi_i = 0.012\text{ Wb}

Reversing the field flips the direction of the flux vectors, resulting in a net change equal to twice the magnitude of the initial flux.

4

Apply Faraday's Law of Electromagnetic Induction to find the induced e.m.f.

\mathcal{E} = N \frac{|\Delta \Phi|}{\Delta t} = 200 \times \frac{0.012}{0.06} = 40\text{ V}

The magnitude of induced e.m.f. equals the total rate of change of magnetic flux linkage across all turns.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction (Magnetic Flux Reversal)

Soru 4397Soru →

A horizontal jet of water issuing from a nozzle with a cross-sectional area of $2.0 \times 10^{-3}\text{ m}^2$ at a speed of $20\text{ m s}^{-1}$ strikes a vertical wall perpendicularly. If the water rebounds horizontally in the opposite direction at a speed of $5.0\text{ m s}^{-1}$ , what is the magnitude of the force exerted by the water stream on the wall? (Density of water = $1000\text{ kg m}^{-3}$ )

600\text{ N}

800\text{ N}

1000\text{ N}

50\text{ N}

Cevabı ve açıklamayı göster

Cevap: $1000\text{ N}$

Cevap

The magnitude of the force exerted by the water jet on the wall is

1000\text{ N}

.

According to Newton's second law, force is the rate of change of linear momentum. The mass of water hitting the wall each second is

\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000 \times 2.0 \times 10^{-3} \times 20 = 40\text{ kg s}^{-1}

. Taking the initial direction as positive (

+20\text{ m s}^{-1}

), the rebounding velocity is opposite in direction (

-5.0\text{ m s}^{-1}

). The magnitude of the change in velocity is

|-5.0 - 20| = 25\text{ m s}^{-1}

. Multiplying mass flow rate by the change in velocity gives a force magnitude of

40 \times 25 = 1000\text{ N}

.

Adım Adım Çözüm

1

Calculate the mass of water striking the wall per second (mass flow rate,

\frac{\Delta m}{\Delta t}

).

\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000\text{ kg m}^{-3} \times (2.0 \times 10^{-3}\text{ m}^2) \times 20\text{ m s}^{-1} = 40\text{ kg s}^{-1}

.

The volume of water reaching the wall per second is given by the cross-sectional area multiplied by its initial speed.

2

Determine the change in velocity vector per unit mass of water

(\Delta v)

.

Taking the direction towards the wall as positive,

v_1 = +20\text{ m s}^{-1}

and

v_2 = -5.0\text{ m s}^{-1}

. Thus,

\Delta v = v_2 - v_1 = -5.0 - 20 = -25.0\text{ m s}^{-1}

.

Velocity is a vector quantity; rebounding in the opposite direction requires assigning opposite signs to the initial and final velocities.

3

Apply Newton's Second Law (

F = \frac{\Delta p}{\Delta t}

) to find the magnitude of force on the wall.

F = \frac{\Delta m}{\Delta t} |\Delta v| = 40\text{ kg s}^{-1} \times 25.0\text{ m s}^{-1} = 1000\text{ N}

.

By Newton's third law, the magnitude of force exerted on the wall equals the rate of change of momentum of the water stream.

Anahtar Kavram

Newton's Second Law and Linear Momentum Rate of Change

Soru 4398Soru →

A solid block of mass $0.60\text{ kg}$ and density $600\text{ kg/m}^3$ is held fully submerged in water of density $1000\text{ kg/m}^3$ by a light vertical string attached to the bottom of a container. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

10.0\text{ N}

4.0\text{ N}

6.0\text{ N}

16.0\text{ N}

Cevabı ve açıklamayı göster

Cevap: $4.0\text{ N}$

Cevap

The tension in the string is

4.0\text{ N}

.

First, the volume of the block is computed as

V = \frac{m}{\rho} = \frac{0.60}{600} = 1.0 \times 10^{-3}\text{ m}^3

. According to Archimedes' principle, the upthrust exerted by the displaced water is

U = \rho_{\text{water}} V g = 1000 \times 1.0 \times 10^{-3} \times 10 = 10.0\text{ N}

. The weight of the block is

W = mg = 0.60 \times 10 = 6.0\text{ N}

. For the block to remain completely submerged in equilibrium, the upward upthrust must balance the downward forces (the weight of the block and the tension

T

pulling downward). Thus,

T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

.

Adım Adım Çözüm

1

Calculate the volume of the block using its mass and density.

V = \frac{m}{\rho_{\text{block}}} = \frac{0.60\text{ kg}}{600\text{ kg/m}^3} = 1.0 \times 10^{-3}\text{ m}^3

The volume of fluid displaced equals the total volume of the fully submerged block.

2

Calculate the upward upthrust force exerted by the water.

U = \rho_{\text{water}} \cdot V \cdot g = 1000\text{ kg/m}^3 \times 1.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 10.0\text{ N}

Archimedes' principle states upthrust equals the weight of the displaced fluid.

3

Calculate the downward gravitational weight of the block.

W = m \cdot g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force exerted on the mass of the block by gravity.

4

Apply vertical force equilibrium to solve for string tension.

U = W + T \Rightarrow T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

The string is tied to the bottom, so tension acts downward to hold the buoyant block in equilibrium.

Anahtar Kavram

Archimedes' Principle and Static Equilibrium of Submerged Bodies

Soru 4399Soru →

A ray of light traveling inside a glass prism of refractive index $1.60$ strikes the boundary with an adjacent oil medium of refractive index $1.20$ . What is the sine of the critical angle for total internal reflection at this glass-oil interface?

0.75

1.33

0.625

0.80

Cevabı ve açıklamayı göster

Cevap: $0.75$

Cevap

The sine of the critical angle at the glass-oil interface is

0.75

.

For light traveling from a medium of refractive index

n_1

into a medium of refractive index

n_2

(where

n_1 > n_2

), the critical angle

C

satisfies

\sin C = \frac{n_2}{n_1}

. Substituting

n_1 = 1.60

and

n_2 = 1.20

gives

\sin C = \frac{1.20}{1.60} = 0.75

.

Adım Adım Çözüm

1

Identify the refractive indices of the two media.

Denser medium (glass):

n_1 = 1.60

; less dense medium (oil):

n_2 = 1.20

.

Total internal reflection occurs when light originates in the optically denser medium and strikes the boundary with a less dense medium.

2

Apply Snell's law at the critical angle

C

.

n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}

.

At the critical angle of incidence, the angle of refraction in the second medium is

90^\circ

, so

\sin 90^\circ = 1

.

3

Substitute the given numerical values to compute

\sin C

.

\sin C = \frac{1.20}{1.60} = \frac{3}{4} = 0.75

.

Dividing the refractive index of the oil by that of the glass gives the exact sine of the critical angle.

Anahtar Kavram

Critical Angle between Two Media

Soru 4400Soru →

A pipe of length $0.80\text{ m}$ is closed at one end and open at the other. If the speed of sound in air is $320\text{ m/s}$ , what is the frequency of its first overtone?

100\text{ Hz}

200\text{ Hz}

300\text{ Hz}

400\text{ Hz}

Cevabı ve açıklamayı göster

Cevap: $300\text{ Hz}$

Cevap

The frequency of the first overtone is

300\text{ Hz}

.

For an air column closed at one end, standing wave resonance occurs only at odd harmonic frequencies given by

f_n = \frac{n v}{4L}

for

n = 1, 3, 5, \dots

. The fundamental frequency (

n = 1

) is

f_1 = \frac{320}{4 \times 0.80} = 100\text{ Hz}

. The first overtone is the very next resonant mode, which corresponds to the third harmonic (

n = 3

), giving

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

.

Adım Adım Çözüm

1

Calculate the fundamental frequency of the closed pipe

f_1 = \frac{v}{4L} = \frac{320}{4 \times 0.80} = 100\text{ Hz}

For a pipe closed at one end, the fundamental wavelength is

\lambda_1 = 4L

.

2

Determine the harmonic number for the first overtone

First overtone = 3rd harmonic (

f_3 = 3 f_1

)

A pipe closed at one end produces only odd harmonics (

f_n = n f_1

where

n = 1, 3, 5, \dots

).

3

Compute the first overtone frequency

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

Multiplying the fundamental frequency by 3 yields the first overtone frequency.

Anahtar Kavram

Resonance and Harmonics in Closed Air Columns

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