Tüm alıştırma soruları

Soru 921Soru →

A international conference delegation of $6$ members is to be selected from $5$ diplomats and $4$ translators. In how many distinct ways can the delegation be formed if it must include at least $4$ diplomats?

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Cevap: 34

Cevap

The total number of distinct ways to form the delegation is 34.

To form a delegation of 6 with at least 4 diplomats out of 5 diplomats and 4 translators, we evaluate two mutually exclusive cases: selecting 4 diplomats and 2 translators (5C4 * 4C2 = 30 ways) and selecting 5 diplomats and 1 translator (5C5 * 4C1 = 4 ways). Summing these gives 34 distinct ways.

Adım Adım Çözüm

1

Determine the valid combinations of diplomats and translators.

Two valid cases exist: (4 diplomats, 2 translators) and (5 diplomats, 1 translator).

The total delegation size is 6 and it must contain at least 4 diplomats out of the 5 available.

2

Compute the selection ways for Case 1 (4 diplomats and 2 translators).

5 * 6 = 30 ways

Selecting 4 diplomats out of 5 is 5C4 = 5 ways, and selecting 2 translators out of 4 is 4C2 = 6 ways.

3

Compute the selection ways for Case 2 (5 diplomats and 1 translator).

1 * 4 = 4 ways

Selecting 5 diplomats out of 5 is 5C5 = 1 way, and selecting 1 translator out of 4 is 4C1 = 4 ways.

4

Sum the combinations from both mutually exclusive cases.

30 + 4 = 34 ways

By the addition principle of counting, mutually exclusive scenarios are added together.

Anahtar Kavram

Combinations with constraints and mutually exclusive cases

Soru 922Soru →

A wooden block of mass $4.0\text{ kg}$ is suspended vertically at rest. A bullet of mass $0.05\text{ kg}$ travelling horizontally at $400\text{ m s}^{-1}$ strikes the block, passes completely through it, and emerges on the opposite side with a reduced speed of $100\text{ m s}^{-1}$ . If a constant retarding force brings the moving block to rest in $0.25\text{ s}$ after the bullet emerges, calculate the magnitude of this retarding force in newtons.

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Cevap: 60

Cevap

The magnitude of the retarding force acting on the block is

60\text{ N}

.

During the impact, the bullet loses momentum equal to

\Delta p = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

. By the conservation of linear momentum, this exact amount of momentum is gained by the block. Applying Newton's second law (

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

), the magnitude of the constant retarding force needed to reduce the block's momentum to zero in

0.25\text{ s}

is

F = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

.

Adım Adım Çözüm

1

Calculate the momentum lost by the bullet during penetration.

\Delta p_{\text{bullet}} = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

The momentum lost by the bullet equals its mass multiplied by the change in its horizontal velocity vector.

2

Determine the initial momentum imparted to the wooden block using the law of conservation of linear momentum.

p_{\text{block}} = \Delta p_{\text{bullet}} = 15\text{ N s}

Since no external horizontal force acts during the collision impact, the momentum lost by the bullet is fully transferred to the block.

3

Calculate the retarding force required to bring the block to rest using the impulse-momentum theorem.

F = \frac{\Delta p_{\text{block}}}{\Delta t} = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

According to Newton's second law of motion, the net force acting on a body equals the rate of change of momentum.

Anahtar Kavram

Conservation of Linear Momentum and Newton's Second Law

Soru 923Soru →

A binary operation $\star$ defined on the set of real numbers $\mathbb{R}$ is given by $x \star y = \frac{x + y}{x - y + 1}$ for $x - y \neq -1$ . If $5 \star p = 3$ , what is the value of $p$ ?

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Cevap: 3.25

Cevap

The value of

p

is

3.25

(or

\frac{13}{4}

).

Applying the operation rule

x \star y = \frac{x + y}{x - y + 1}

with

x = 5

and

y = p

gives

\frac{5 + p}{6 - p} = 3

. Cross-multiplying gives

5 + p = 18 - 3p

, which simplifies to

4p = 13

or

p = 3.25

.

Adım Adım Çözüm

1

Substitute

x = 5

and

y = p

into the given binary operation definition

\frac{5 + p}{5 - p + 1} = 3

This sets up the equation for the given condition

5 \star p = 3

.

2

Simplify the denominator in the algebraic fraction

\frac{5 + p}{6 - p} = 3

Combining the constants

5 + 1 = 6

simplifies the denominator.

3

Multiply both sides by

(6 - p)

and expand

5 + p = 18 - 3p

Eliminating the denominator allows linear terms in

p

to be collected.

4

Rearrange terms to solve for

p

4p = 13 \implies p = 3.25

Adding

3p

to both sides and subtracting

5

gives

4p = 13

.

Anahtar Kavram

Solving linear equations derived from non-commutative binary operations

Soru 924Soru →

An electric crane lifts a load of $250\text{ kg}$ vertically upwards through a height of $12\text{ m}$ in $10\text{ s}$ at a constant speed. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the useful output power of the crane in watts?

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Cevap: 3000

Cevap

The useful output power of the crane is

3000\text{ W}

.

The work done in lifting the load vertically is equal to the gravitational potential energy gained,

W = mgh = 250 \times 10 \times 12 = 30,000\text{ J}

. Power is the rate of doing work, so

P = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}

.

Adım Adım Çözüm

1

Identify the given values

Mass

m = 250\text{ kg}

, height

h = 12\text{ m}

, time

t = 10\text{ s}

, acceleration due to gravity

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

.

Extract values needed for work and power calculations.

2

Calculate the work done in lifting the load

W = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}

.

The work done against gravity equals the gain in gravitational potential energy.

3

Calculate the power output

P = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}

.

Power is defined as the rate at which work is done (

P = \frac{W}{t}

).

Anahtar Kavram

Power as the rate of doing work against gravity

Soru 925Soru →

The volume flow rate $Q$ of a viscous liquid through a pipe depends on the radius $r$ of the pipe, the coefficient of viscosity $\eta$ , and the pressure gradient $\frac{\Delta P}{L}$ according to the dimensional equation $Q = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z$ , where $k$ is a dimensionless constant. What is the value of the exponent $x$ ?

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Cevap: 4

Cevap

The value of the exponent

x

is 4.

Applying the principle of dimensional homogeneity, the dimensions of volume flow rate

[Q] = L^3 T^{-1}

are equated to

[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z

. Equating powers yields

y + z = 0

for mass,

-y - 2z = -1

for time, and

x - y - 2z = 3

for length. Solving these simultaneous equations gives

z = 1

,

y = -1

, and

x = 4

.

Adım Adım Çözüm

1

Identify the base dimensions of each physical quantity in the given equation.

Flow rate

[Q] = M^0 L^3 T^{-1}

, radius

[r] = L

, viscosity

[\eta] = M L^{-1} T^{-1}

, and pressure gradient

\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}

.

Expressing each quantity in terms of fundamental dimensions (

M

,

L

,

T

) is necessary for dimensional analysis.

2

Substitute dimensions into the power-law equation and collect powers of base dimensions.

M^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}

.

The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.

3

Set up and solve linear equations for the exponents

x

,

y

, and

z

.

Solving

y + z = 0

,

-y - 2z = -1

, and

x - y - 2z = 3

yields

z = 1

,

y = -1

, and

x = 4

.

Equating powers of

M

,

T

, and

L

allows step-by-step determination of each unknown exponent.

Anahtar Kavram

Dimensions of Physical Quantities and Dimensional Analysis

Soru 926Soru →

A body of mass $2.5\text{ kg}$ moving at a speed of $4.0\text{ m s}^{-1}$ along a straight horizontal path is brought to rest in $2.0\text{ s}$ by a constant retarding force. What is the magnitude of this retarding force in newtons?

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Cevap: 5

Cevap

The magnitude of the retarding force is

5.0\text{ N}

.

According to Newton's second law of motion, net force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}

). Substituting

m = 2.5\text{ kg}

,

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

,

v = 0\text{ m s}^{-1}

, and

t = 2.0\text{ s}

gives

F = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}

. The magnitude of this force is

5.0\text{ N}

.

Adım Adım Çözüm

1

Determine the initial momentum and final momentum of the body.

Initial momentum

p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}

, and final momentum

p_f = 0\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity (

p = mv

).

2

Calculate the magnitude of the force applied using the impulse-momentum relationship

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

.

Magnitude of force

F = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}

.

Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Anahtar Kavram

Newton's Second Law and Impulse-Momentum Relationship

Tahmini Süre:45s

Soru 927Soru →

Monochromatic radiation carrying photons of energy $4.8\text{ eV}$ illuminates a cesium surface inside a photoelectric cell. If the work function of cesium is $2.1\text{ eV}$ , determine the stopping potential, in volts, needed to reduce the photoelectric current to zero.

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Cevap: 2.7

Cevap

The stopping potential required to reduce the photoelectric current to zero is

2.7\text{ V}

.

Einstein's photoelectric equation states that incident photon energy

E

equals the work function

W_0

plus the maximum kinetic energy

K_{\text{max}}

of the photoelectrons (

E = W_0 + K_{\text{max}}

). Rearranging gives

K_{\text{max}} = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

. Since

K_{\text{max}} = e V_s

, an electron-volt value of kinetic energy numerically equals the stopping potential in volts, giving a stopping potential of

2.7\text{ V}

.

Adım Adım Çözüm

1

Calculate the maximum kinetic energy (

K_{\text{max}}

) of the emitted photoelectrons.

K_{\text{max}} = E - W_0 = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

According to Einstein's photoelectric equation, incident photon energy is divided into overcoming the work function of the metal and providing kinetic energy to the liberated electron.

2

Determine the stopping potential (

V_s

) from the maximum kinetic energy.

V_s = \frac{K_{\text{max}}}{e} = \frac{2.7\text{ eV}}{e} = 2.7\text{ V}

The stopping potential

V_s

is the opposing potential difference needed to stop the fastest moving photoelectrons, defined by

K_{\text{max}} = e V_s

.

Anahtar Kavram

Photoelectric Effect and Work Function

Soru 928Soru →

An electric cell of electromotive force $E$ and internal resistance $r$ is connected in series with a fixed resistor of resistance $8.0\text{ }\Omega$ and a galvanometer of internal resistance $40.0\text{ }\Omega$ . When a shunt resistor of resistance $10.0\text{ }\Omega$ is connected in parallel across the galvanometer, the total current supplied by the cell increases by $50\%$ . What is the internal resistance $r$ of the cell in ohms?

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Cevap: 48

Cevap

The internal resistance of the cell is

48.0\text{ }\Omega

.

By determining the initial total circuit resistance

R_{T1} = 48.0 + r

and the post-shunt total circuit resistance

R_{T2} = 16.0 + r

, we apply Ohm's law with

I_2 = 1.5 I_1

. Equating total supply voltage

E

gives

\frac{E}{16.0 + r} = 1.5 \times \frac{E}{48.0 + r}

, which simplifies directly to

r = 48.0\text{ }\Omega

.

Adım Adım Çözüm

1

Formulate the initial total resistance of the series circuit

R_{T1} = 8.0\text{ }\Omega + 40.0\text{ }\Omega + r = 48.0 + r

Before the shunt is added, the fixed resistor, galvanometer, and internal resistance of the cell are all connected in series.

2

Calculate the effective resistance of the shunted galvanometer

R_p = \frac{40.0 \times 10.0}{40.0 + 10.0} = 8.0\text{ }\Omega

Connecting the shunt in parallel across the galvanometer forms a parallel network.

3

Formulate the final total circuit resistance after shunting

R_{T2} = 8.0\text{ }\Omega + 8.0\text{ }\Omega + r = 16.0 + r

The new circuit consists of the fixed resistor, the parallel shunted galvanometer combination, and the internal resistance in series.

4

Set up the ratio equation for total circuit currents based on the given

50\%

current increase

\frac{E}{16.0 + r} = 1.5 \left(\frac{E}{48.0 + r}\right)

An increase of

50\%

means

I_2 = 1.5 I_1

. According to Ohm's law, total current is inversely proportional to total circuit resistance for a constant EMF

E

.

5

Solve the algebraic equation for internal resistance

r

r = 48.0\text{ }\Omega

Cross-multiplying yields

48.0 + r = 24.0 + 1.5r

, leading directly to

0.5r = 24.0

, so

r = 48.0\text{ }\Omega

.

Anahtar Kavram

Internal Resistance and Circuit Modification via Galvanometer Shunting

Soru 929Soru →

In a manufacturing plant, two independent automated assembly units, $U_1$ and $U_2$ , undergo safety inspection. The probability that unit $U_1$ fails the inspection is $\frac{1}{5}$ , while the probability that unit $U_2$ fails the inspection is $\frac{1}{4}$ . What is the probability that at least one of the two units fails the inspection? Express your answer as a decimal.

Cevabı ve açıklamayı göster

Cevap: 0.4

Cevap

0.4

The probability that at least one unit fails is determined using the addition law for non-mutually exclusive independent events:

P(U_1 \cup U_2) = P(U_1) + P(U_2) - P(U_1 \cap U_2) = 0.20 + 0.25 - (0.20 \times 0.25) = 0.45 - 0.05 = 0.40

. Alternatively, using the complement rule gives

1 - P(\text{neither fails}) = 1 - (1 - 0.20)(1 - 0.25) = 1 - 0.60 = 0.40

.

Adım Adım Çözüm

1

Identify the individual failure probabilities

P(U_1) = 0.20

and

P(U_2) = 0.25

These values represent the single event failure probabilities given in the problem statement.

2

Calculate the probability that neither unit fails

P(U_1') \times P(U_2') = (1 - 0.20) \times (1 - 0.25) = 0.80 \times 0.75 = 0.60

Because the units operate independently, their complement events (passing inspection) are also independent.

3

Determine the probability of at least one unit failing

1 - 0.60 = 0.40

The event that at least one unit fails is the complementary event of neither unit failing.

Anahtar Kavram

Addition and Multiplication Laws of Probability for Independent Compound Events

Soru 930Soru →

A person standing at a stationary position between two tall parallel vertical walls fires a starter pistol. The person hears the first echo reflected from the nearer wall after $1.2\text{ s}$ and the second echo from the farther wall after $1.8\text{ s}$ . Given that the speed of sound in air is $340\text{ m/s}$ , what is the total distance between the two walls in meters?

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Cevap: 510

Cevap

The total distance between the two walls is

510\text{ m}

.

Because sound travels to each wall and reflects back to the observer, the distance to each wall is given by

d = \frac{v \cdot t}{2}

. The distance to the nearer wall is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

, and the distance to the farther wall is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. Since the observer is between the two walls, the total separation between the walls is

204\text{ m} + 306\text{ m} = 510\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance from the observer to the nearer wall.

d_1 = 204\text{ m}

The sound travels to the nearer wall and back in

1.2\text{ s}

, covering twice the distance to that wall.

2

Calculate the distance from the observer to the farther wall.

d_2 = 306\text{ m}

The sound travels to the farther wall and back in

1.8\text{ s}

, covering twice the distance to that wall.

3

Add the two individual distances to find the total distance between the walls.

D = d_1 + d_2 = 510\text{ m}

The observer is positioned between the two walls, so the separation distance is the sum of both distances.

Anahtar Kavram

Echo and Speed of Sound Propagation between Parallel Boundaries

Soru 931Soru →

The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = 2n^2 + 3n$ . What is the $8^{\text{th}}$ term of the progression?

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Cevap: 33

Cevap

The

8^{\text{th}}

term of the arithmetic progression is

33

.

The

n^{\text{th}}

term of any sequence can be calculated using the identity

T_n = S_n - S_{n-1}

. Substituting

n = 8

gives

S_8 = 2(8)^2 + 3(8) = 152

, and substituting

n = 7

gives

S_7 = 2(7)^2 + 3(7) = 119

. Therefore,

T_8 = 152 - 119 = 33

.

Adım Adım Çözüm

1

Calculate the sum of the first 8 terms (

S_8

) by substituting

n = 8

into

S_n = 2n^2 + 3n

.

S_8 = 2(8)^2 + 3(8) = 128 + 24 = 152

.

The sum formula provides the total sum of terms from

T_1

to

T_8

.

2

Calculate the sum of the first 7 terms (

S_7

) by substituting

n = 7

into

S_n = 2n^2 + 3n

.

S_7 = 2(7)^2 + 3(7) = 98 + 21 = 119

.

The sum formula provides the total sum of terms from

T_1

to

T_7

.

3

Subtract

S_7

from

S_8

to determine the value of the

8^{\text{th}}

term (

T_8

).

T_8 = S_8 - S_7 = 152 - 119 = 33

.

The difference between the sum of the first

n

terms and the sum of the first

(n-1)

terms yields the

n^{\text{th}}

term (

T_n = S_n - S_{n-1}

).

Anahtar Kavram

Finding the nth term of a sequence from the sum of the first n terms using

T_n = S_n - S_{n-1}

.

Tahmini Süre:1m 30s

Soru 932Soru →

A rectangular glass block of thickness $6.0\text{ cm}$ has a refractive index of $1.50$ . Calculate the apparent depth, in centimeters, of a mark placed at the bottom of the block when viewed normally from above.

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The apparent depth of the mark is

4.0\text{ cm}

.

The refractive index of a medium relative to air is given by the ratio of real depth to apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Substituting the given values gives

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

.

Adım Adım Çözüm

1

Identify the relationship between refractive index, real depth, and apparent depth for normal view

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

By definition of optical refraction when looking normally from an optically less dense medium (air) into a denser medium (glass).

2

Substitute the known values (

n = 1.50

,

\text{Real Depth} = 6.0\text{ cm}

) and solve for the apparent depth

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

Dividing the real thickness by the refractive index yields the perceived (apparent) depth.

Anahtar Kavram

Real and Apparent Depth in Refraction

Soru 933Soru →

An electric train traveling along a straight track uniformly slows down from a speed of $30\text{ m/s}$ to $10\text{ m/s}$ over a distance of $200\text{ m}$ . What is the magnitude of the deceleration of the train in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The magnitude of the deceleration is

2\text{ m/s}^2

.

Using the kinematic equation

v^2 = u^2 + 2as

, substituting

v = 10\text{ m/s}

,

u = 30\text{ m/s}

, and

s = 200\text{ m}

gives

100 = 900 + 400a

, which simplifies to

400a = -800

, yielding

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

. The magnitude of deceleration is

2\text{ m/s}^2

.

Adım Adım Çözüm

1

Identify the given kinematic variables.

Initial velocity

u = 30\text{ m/s}

, final velocity

v = 10\text{ m/s}

, displacement

s = 200\text{ m}

.

Choosing the appropriate equation of motion requires knowing which variables are given and which is unknown.

2

Apply the third equation of motion relating initial velocity, final velocity, acceleration, and distance.

v^2 = u^2 + 2as

This formula connects

u

,

v

,

a

, and

s

without needing time

t

.

3

Substitute the given values into the equation and solve for acceleration

a

.

(10)^2 = (30)^2 + 2(a)(200) \implies 100 = 900 + 400a \implies 400a = -800 \implies a = -2\text{ m/s}^2

.

Performing algebraic operations to isolate the acceleration parameter.

4

State the magnitude of the deceleration.

The magnitude of deceleration is

2\text{ m/s}^2

.

Deceleration represents the rate of speed reduction, which corresponds to the magnitude of negative acceleration.

Anahtar Kavram

Uniformly Accelerated Motion Equations

Soru 934Soru →

A proton of mass $1.67 \times 10^{-27}\text{ kg}$ and an electron of mass $9.11 \times 10^{-31}\text{ kg}$ , both carrying charges of equal magnitude, are accelerated from rest through the same electric potential difference. Calculate the ratio of the de Broglie wavelength of the electron to that of the proton.

Cevabı ve açıklamayı göster

Cevap: 42.8

Cevap

The ratio of the de Broglie wavelength of the electron to that of the proton is 42.8.

The de Broglie wavelength of a particle accelerated through potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2mqV}}

. Since both particles carry equal charge

q

and experience the same potential

V

, the ratio of their wavelengths simplifies to

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

.

Adım Adım Çözüm

1

Relate kinetic energy to accelerating potential difference

E_k = qV

Electric potential energy converted into kinetic energy during acceleration from rest.

2

Express de Broglie wavelength in terms of particle mass, charge, and potential difference

\lambda = \frac{h}{\sqrt{2mqV}}

Combining

p = \sqrt{2mE_k}

with de Broglie's formula

\lambda = \frac{h}{p}

.

3

Formulate the wavelength ratio of electron to proton

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}

Planck's constant

h

, elementary charge

q

, and potential difference

V

are identical for both particles and cancel out.

4

Substitute numerical values and compute final ratio

\sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

Square root of the proton-to-electron mass ratio yields the inverse ratio of their wavelengths.

Anahtar Kavram

De Broglie wavelength relation to particle mass under constant accelerating potential

Soru 935Soru →

Two fishing boats leave a harbor $H$ at the same time. Boat $A$ travels on a bearing of $020^\circ$ for a distance of $8\text{ km}$ , while Boat $B$ travels on a bearing of $140^\circ$ for a distance of $7\text{ km}$ . What is the distance between Boat $A$ and Boat $B$ in kilometers?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

13 km

The distance between the two boats forms the third side of triangle

HAB

, where

HA = 8\text{ km}

,

HB = 7\text{ km}

, and the included angle at the harbor

H

is

\angle AHB = 140^\circ - 020^\circ = 120^\circ

. By the Cosine Rule,

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

. Taking the square root gives

AB = 13\text{ km}

.

Adım Adım Çözüm

1

Calculate the included angle between the direction vectors of the two boats from the harbor.

Included angle

\angle AHB = 140^\circ - 020^\circ = 120^\circ

The angle between two bearings originating from the same point is the difference between their bearing angles.

2

State the Cosine Rule for side

AB

in triangle

HAB

.

AB^2 = HA^2 + HB^2 - 2 \cdot HA \cdot HB \cdot \cos(\angle AHB)

The Cosine Rule calculates an unknown side when two sides and their included angle (SAS) are given.

3

Substitute given side lengths

HA = 8\text{ km}

,

HB = 7\text{ km}

, and angle

\angle AHB = 120^\circ

.

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

Since

120^\circ

is in the second quadrant,

\cos(120^\circ) = -0.5

, which changes the subtracted term to addition.

4

Compute the principal square root of 169.

AB = \sqrt{169} = 13\text{ km}

Distance is a non-negative scalar quantity.

Anahtar Kavram

Applying the Cosine Rule to solve bearing problems

Soru 936Soru →

A simple pendulum completes $20$ full oscillations in a time interval of $40\text{ s}$ . What is the period of oscillation of the pendulum in seconds?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The period of oscillation of the pendulum is

2.0\text{ s}

.

The period

T

of a repeating motion is the time taken to complete one full cycle. Dividing the total measured time (

40\text{ s}

) by the number of oscillations (

20

) gives

2.0\text{ s}

.

Adım Adım Çözüm

1

Identify the given total time and total number of oscillations

Total time

t = 40\text{ s}

, number of oscillations

N = 20

The period is defined as the time taken for one single complete oscillation.

2

Apply the period formula

T = \frac{t}{N}

T = \frac{40\text{ s}}{20} = 2.0\text{ s}

Dividing the total measured time by the total count of oscillations yields the time per oscillation.

Anahtar Kavram

Period of a Simple Pendulum

Soru 937Soru →

A gas sample enclosed in a rigid container of fixed volume has a root-mean-square (r.m.s.) speed of $500\text{ m s}^{-1}$ at a temperature of $127^\circ\text{C}$ . If the gas is heated until its pressure is quadrupled, what is the new r.m.s. speed of the gas molecules in $\text{m s}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 1000

Cevap

1000

According to kinetic theory, the pressure of a fixed volume of gas is directly proportional to its absolute temperature (

P \propto T

), and the root-mean-square speed of its molecules is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). Quadrupling the pressure quadruples the absolute temperature in Kelvin from

400\text{ K}

to

1600\text{ K}

. Since the speed scales as

\sqrt{4} = 2

, the initial r.m.s. speed of

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

doubles to

1000\text{ m s}^{-1}

.

Adım Adım Çözüm

1

Convert the initial temperature to absolute temperature (Kelvin)

T_1 = 127^\circ\text{C} + 273 = 400\text{ K}

Kinetic theory relationships and gas laws require temperature in absolute units (Kelvin).

2

Determine the new absolute temperature based on the pressure change at constant volume

T_2 = 4 \times T_1 = 1600\text{ K}

For a fixed volume of gas, pressure is directly proportional to absolute temperature (

P \propto T

). Therefore, quadrupling the pressure quadruples the absolute temperature.

3

Calculate the new root-mean-square speed using the square root relationship

v_2 = v_1 \sqrt{\frac{T_2}{T_1}} = 500 \times \sqrt{4} = 1000\text{ m s}^{-1}

Root-mean-square speed is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Anahtar Kavram

Relationship between microscopic kinetic parameters (r.m.s. speed) and macroscopic state variables (pressure and absolute temperature)

Soru 938Soru →

A cell of electromotive force $E = 6.0\text{ V}$ and internal resistance $r = 1.0\text{ }\Omega$ is connected in series with a resistor $R$ and a shunted galvanometer. The galvanometer has a resistance of $90\text{ }\Omega$ and produces a full-scale deflection for a current of $2.0\text{ mA}$ . If the shunt resistance connected across the galvanometer is $10\text{ }\Omega$ , calculate the value of the series resistor $R$ , in ohms $(\Omega)$ , required for the galvanometer to show full-scale deflection.

Cevabı ve açıklamayı göster

Cevap: 290

Cevap

The required value of the series resistor

R

is

290\text{ }\Omega

.

At full-scale deflection, a current of

2.0\text{ mA}

passes through the galvanometer, resulting in a potential difference of

V_g = 2.0 \times 10^{-3} \times 90 = 0.18\text{ V}

. Since the shunt is connected in parallel with the galvanometer, the current through the shunt is

I_s = \frac{0.18}{10} = 0.018\text{ A} = 18\text{ mA}

. Thus, the total current supplied by the cell is

I = 2\text{ mA} + 18\text{ mA} = 20\text{ mA} = 0.02\text{ A}

. The total equivalent resistance of the shunted galvanometer is

R_p = \frac{90 \times 10}{90 + 10} = 9.0\text{ }\Omega

. Applying Ohm's law to the total loop including internal resistance

r

, we have

E = I(R + R_p + r) \implies 6.0 = 0.02(R + 9.0 + 1.0) \implies R + 10.0 = 300 \implies R = 290\text{ }\Omega

.

Adım Adım Çözüm

1

Calculate voltage across the galvanometer at full-scale deflection

V_g = 0.18\text{ V}

The potential difference across parallel branches is equal, and for full-scale deflection

I_g = 2.0\text{ mA}

.

2

Calculate the current passing through the shunt resistor

I_s = 18.0\text{ mA} = 0.018\text{ A}

Using Ohm's law across the shunt resistor

S = 10\text{ }\Omega

with

V_s = V_g = 0.18\text{ V}

.

3

Calculate the total circuit current provided by the cell

I = 20.0\text{ mA} = 0.020\text{ A}

By Kirchhoff's current law, the main current splits between the galvanometer and shunt.

4

Find the equivalent resistance of the shunted galvanometer and total circuit resistance

R_p = 9.0\text{ }\Omega

and total circuit resistance

R_{\text{total}} = 300\text{ }\Omega

Parallel resistance formula gives

R_p = 9.0\text{ }\Omega

, and

R_{\text{total}} = \frac{E}{I} = \frac{6.0\text{ V}}{0.020\text{ A}} = 300\text{ }\Omega

.

5

Solve for the unknown external series resistance

R

R = 290\text{ }\Omega

R_{\text{total}} = R + r + R_p \implies 300 = R + 1.0 + 9.0 \implies R = 290\text{ }\Omega

.

Anahtar Kavram

Galvanometer Shunting and Electric Circuit Analysis with Internal Resistance

Soru 939Soru →

A $24.6\text{ g}$ sample of hydrated magnesium tetraoxosulfate(VI), $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ , was heated strongly until all the water of crystallization was driven off, leaving behind $12.0\text{ g}$ of anhydrous $\text{MgSO}_4$ . What is the value of $x$ ? $[\text{Mg} = 24, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

x

is 7.

Heating the sample evaporates

12.6\text{ g}

of water of crystallization (

24.6\text{ g} - 12.0\text{ g}

). Converting the masses to moles gives

0.1\text{ mol}

of anhydrous

\text{MgSO}_4

(

12.0\text{ g} / 120\text{ g/mol}

) and

0.7\text{ mol}

of

\text{H}_2\text{O}

(

12.6\text{ g} / 18\text{ g/mol}

). The mole ratio of water to salt is

0.7 / 0.1 = 7

, which means

x = 7

.

Adım Adım Çözüm

1

Calculate the mass of water of crystallization lost during heating.

Mass of water =

24.6\text{ g} - 12.0\text{ g} = 12.6\text{ g}

.

The decrease in mass after heating represents the mass of water driven off from the hydrated salt.

2

Determine the molar masses of

\text{MgSO}_4

and

\text{H}_2\text{O}

.

Molar mass of

\text{MgSO}_4 = 24 + 32 + (4 \times 16) = 120\text{ g/mol}

; Molar mass of

\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}

.

Molar masses are required to convert the measured masses into chemical amounts (moles).

3

Calculate the number of moles of anhydrous salt and water.

Moles of

\text{MgSO}_4 = \frac{12.0\text{ g}}{120\text{ g/mol}} = 0.1\text{ mol}

; Moles of

\text{H}_2\text{O} = \frac{12.6\text{ g}}{18\text{ g/mol}} = 0.7\text{ mol}

.

The chemical formula stoichiometry is determined by the molar ratio of components.

4

Determine the mole ratio of water to anhydrous salt (

x

).

x = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{MgSO}_4} = \frac{0.7\text{ mol}}{0.1\text{ mol}} = 7

.

The coefficient

x

is the integer ratio of moles of water of crystallization per mole of anhydrous salt.

Anahtar Kavram

Water of Crystallization Stoichiometry

Soru 940Soru →

Car $A$ , traveling at a constant speed of $20\text{ m/s}$ along a straight horizontal road, passes a landmark $50\text{ m}$ ahead of car $B$ , which is initially stationary. At $t = 0\text{ s}$ , car $B$ starts moving in the same direction, accelerating uniformly at $3.0\text{ m/s}^2$ until it reaches a top speed of $30\text{ m/s}$ , after which it continues at this constant top speed. How many seconds after $t = 0\text{ s}$ does car $B$ catch up with car $A$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

Car B catches up with car A after 20 seconds.

Car B accelerates for

10\text{ s}

covering

150\text{ m}

to reach

30\text{ m/s}

. During these

10\text{ s}

, car A reaches a position of

250\text{ m}

(taking into account its

50\text{ m}

head start). Car B then closes the remaining

100\text{ m}

gap at a relative speed of

10\text{ m/s}

(

30\text{ m/s} - 20\text{ m/s}

), taking an extra

10\text{ s}

. The total elapsed time is

10\text{ s} + 10\text{ s} = 20\text{ s}

.

Adım Adım Çözüm

1

Calculate the duration

t_1

of car B's acceleration phase to reach

30\text{ m/s}

.

t_1 = \frac{v_{max} - u}{a} = \frac{30 - 0}{3.0} = 10\text{ s}

Car B accelerates uniformly from rest at

3.0\text{ m/s}^2

until reaching its top speed limit.

2

Determine the distance

s_B

covered by car B and the position

s_A

of car A at

t = 10\text{ s}

.

s_B = \frac{1}{2}a t_1^2 = \frac{1}{2}(3.0)(10)^2 = 150\text{ m}

;

s_A = 50 + v_A t_1 = 50 + (20)(10) = 250\text{ m}

Car A starts

50\text{ m}

ahead and moves continuously at

20\text{ m/s}

.

3

Find the separation distance between the two cars at

t = 10\text{ s}

and compute the time

\Delta t

required to close it.

\text{Separation} = 250 - 150 = 100\text{ m}; \Delta t = \frac{100}{30 - 20} = 10\text{ s}

Beyond

t = 10\text{ s}

, car B travels at a constant relative velocity of

10\text{ m/s}

faster than car A.

4

Sum the acceleration time and constant speed time to find the total time taken.

t_{total} = t_1 + \Delta t = 10\text{ s} + 10\text{ s} = 20\text{ s}

Combining both phases yields the exact instant car B overtakes car A.

Anahtar Kavram

Multi-stage relative motion with acceleration limits

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