Tüm alıştırma soruları

13931 soru

Soru 4281Soru

If 4+32322\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}} is expressed in the simplified form a+b2a + b\sqrt{2}, where aa and bb are integers, what is the value of a+ba + b?

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Cevap: 41

Cevap

The value of a+ba + b is 41.
To rationalise 4+32322\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}, multiply both numerator and denominator by the conjugate 3+223 + 2\sqrt{2}. The denominator becomes 32(22)2=98=13^2 - (2\sqrt{2})^2 = 9 - 8 = 1. Expanding the numerator gives (4)(3)+4(22)+32(3)+32(22)=12+82+92+12=24+172(4)(3) + 4(2\sqrt{2}) + 3\sqrt{2}(3) + 3\sqrt{2}(2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}. Comparing with a+b2a + b\sqrt{2} gives a=24a = 24 and b=17b = 17, so a+b=24+17=41a + b = 24 + 17 = 41.

Adım Adım Çözüm

1
Multiply the numerator and denominator by the conjugate of the denominator.
The expression becomes (4+32)(3+22)(322)(3+22)\frac{(4 + 3\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}.
Multiplying by the conjugate eliminates surds from the denominator using the difference of two squares identity (xy)(x+y)=x2y2(x-y)(x+y) = x^2 - y^2.
2
Simplify the denominator.
(3)2(22)2=9(4×2)=98=1(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1.
Squaring 222\sqrt{2} yields 22×(2)2=4×2=82^2 \times (\sqrt{2})^2 = 4 \times 2 = 8.
3
Expand the numerator.
(4×3)+(4×22)+(32×3)+(32×22)=12+82+92+12=24+172(4 \times 3) + (4 \times 2\sqrt{2}) + (3\sqrt{2} \times 3) + (3\sqrt{2} \times 2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}.
Applying the distributive law and grouping rational terms together and like surd terms together.
4
Identify the values of aa and bb and calculate a+ba + b.
a=24a = 24, b=17b = 17, so a+b=24+17=41a + b = 24 + 17 = 41.
Matching coefficients of the simplified surd form a+b2a + b\sqrt{2}.

Anahtar Kavram

Rationalisation of binomial surd denominators using conjugates
Tahmini Süre:2m 0s
Soru 4282Soru

The total profit P(x)P(x), in thousands of Naira, obtained from producing and selling xx hundred units of a commodity is modeled by the function P(x)=x3+6x2+15x8P(x) = -x^3 + 6x^2 + 15x - 8, where x0x \ge 0. What is the maximum profit achievable?

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Cevap: 92

Cevap

The maximum profit achievable is 92 thousand Naira.
To find the maximum profit, we find the stationary points of P(x)=x3+6x2+15x8P(x) = -x^3 + 6x^2 + 15x - 8 by taking the derivative P(x)=3x2+12x+15P'(x) = -3x^2 + 12x + 15 and setting it to 0. Solving 3(x5)(x+1)=0-3(x-5)(x+1) = 0 with x0x \ge 0 yields x=5x = 5. Testing the second derivative gives P(5)=18<0P''(5) = -18 < 0, confirming x=5x = 5 is a maximum. Substituting x=5x = 5 into P(x)P(x) yields P(5)=92P(5) = 92.

Adım Adım Çözüm

1
Differentiate the profit function P(x)P(x) with respect to xx
P(x)=3x2+12x+15P'(x) = -3x^2 + 12x + 15
Stationary points occur where the rate of change of profit (the derivative) is equal to zero.
2
Set P(x)=0P'(x) = 0 and solve for xx
3(x24x5)=0    (x5)(x+1)=0-3(x^2 - 4x - 5) = 0 \implies (x - 5)(x + 1) = 0, giving critical values x=5x = 5 and x=1x = -1
Factoring the quadratic equation yields the critical values of production level.
3
Filter critical values based on physical domain constraints
x=5x = 5 (reject x=1x = -1 since production x0x \ge 0)
Production quantities cannot be negative in physical real-life contexts.
4
Perform the second derivative test to confirm the nature of the stationary point
P(x)=6x+12    P(5)=6(5)+12=18P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18
Since P(5)<0P''(5) < 0, the function achieves a local maximum at x=5x = 5.
5
Substitute x=5x = 5 back into original function P(x)P(x)
P(5)=(5)3+6(5)2+15(5)8=125+150+758=92P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92
Evaluating P(5)P(5) gives the maximum total profit value.

Anahtar Kavram

Optimization and Maximum Values using First and Second Derivatives
Soru 4283Soru

If the line 3x+py7=03x + py - 7 = 0 is perpendicular to the line passing through the points (1,2)(1, -2) and (4,7)(4, 7), what is the value of pp?

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Cevap: 99

Cevap

The value of pp is 99.
The line passing through (1,2)(1, -2) and (4,7)(4, 7) has a gradient of m1=7(2)41=3m_1 = \frac{7 - (-2)}{4 - 1} = 3. The equation 3x+py7=03x + py - 7 = 0 can be rewritten as y=3px+7py = -\frac{3}{p}x + \frac{7}{p}, giving a gradient of m2=3pm_2 = -\frac{3}{p}. For perpendicular lines, the product of their gradients must equal 1-1, so 3×(3p)=13 \times \left(-\frac{3}{p}\right) = -1, which simplifies to p=9p = 9.

Adım Adım Çözüm

1
Calculate the gradient (m1m_1) of the line passing through (1,2)(1, -2) and (4,7)(4, 7)
m1=7(2)41=93=3m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3
The gradient between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Express the line 3x+py7=03x + py - 7 = 0 in slope-intercept form (y=mx+cy = mx + c) to find its gradient (m2m_2)
py=3x+7    y=3px+7ppy = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}, so m2=3pm_2 = -\frac{3}{p}
The coefficient of xx when solved for yy represents the gradient of the straight line.
3
Apply the perpendicularity condition m1m2=1m_1 \cdot m_2 = -1 and solve for pp
3(3p)=1    9p=1    p=93 \cdot \left(-\frac{3}{p}\right) = -1 \implies -\frac{9}{p} = -1 \implies p = 9
Two non-vertical lines are perpendicular if and only if the product of their gradients is 1-1.

Anahtar Kavram

Perpendicular Lines and Gradients
Tahmini Süre:1m 30s
Soru 4284Soru

What is the value of the definite integral 13(2x+1)dx\int_{1}^{3} (2x + 1) \, dx?

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Cevap: 10

Cevap

10
The antiderivative of 2x+12x + 1 is x2+xx^2 + x. Evaluating at the upper boundary x=3x = 3 gives 32+3=123^2 + 3 = 12, and at the lower boundary x=1x = 1 gives 12+1=21^2 + 1 = 2. Subtracting the lower bound result from the upper bound result yields 122=1012 - 2 = 10.

Adım Adım Çözüm

1
Find the indefinite integral (antiderivative) of 2x+12x + 1
(2x+1)dx=x2+x\int (2x + 1) \, dx = x^2 + x
Apply the power rule of integration: xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
2
Evaluate the antiderivative at the upper limit x=3x = 3
32+3=9+3=123^2 + 3 = 9 + 3 = 12
Substitute x=3x = 3 into x2+xx^2 + x.
3
Evaluate the antiderivative at the lower limit x=1x = 1
12+1=1+1=21^2 + 1 = 1 + 1 = 2
Substitute x=1x = 1 into x2+xx^2 + x.
4
Subtract the lower limit value from the upper limit value
122=1012 - 2 = 10
According to the Fundamental Theorem of Calculus: abf(x)dx=F(b)F(a)\int_{a}^{b} f(x)dx = F(b) - F(a).

Anahtar Kavram

Definite Integrals of Linear Polynomials
Soru 4285Soru

An object of mass 0.20 kg0.20\text{ kg} executes simple harmonic motion with an amplitude of 0.05 m0.05\text{ m} and a maximum acceleration of 20 m/s220\text{ m/s}^2. What is the speed of the object when its displacement from the equilibrium position is 0.03 m0.03\text{ m}?

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Cevap: 0.80 m/s0.80\text{ m/s}

Cevap

The speed of the object at a displacement of 0.03 m0.03\text{ m} is 0.80 m/s0.80\text{ m/s}.
The maximum acceleration in simple harmonic motion is given by amax=ω2Aa_{\text{max}} = \omega^2 A. Substituting amax=20 m/s2a_{\text{max}} = 20\text{ m/s}^2 and A=0.05 mA = 0.05\text{ m} gives ω2=400 rad2/s2\omega^2 = 400\text{ rad}^2/\text{s}^2, so ω=20 rad/s\omega = 20\text{ rad/s}. The speed at any displacement xx is given by v=ωA2x2v = \omega \sqrt{A^2 - x^2}. For x=0.03 mx = 0.03\text{ m}, v=200.0520.032=20×0.04=0.80 m/sv = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}.

Adım Adım Çözüm

1
Determine the angular frequency (ω\omega) of the simple harmonic motion from the maximum acceleration formula.
ω=20 rad/s\omega = 20\text{ rad/s}
Maximum acceleration is given by amax=ω2Aa_{\text{max}} = \omega^2 A. Rearranging gives ω2=amaxA=200.05=400 rad2/s2\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2, so ω=20 rad/s\omega = 20\text{ rad/s}.
2
Calculate the speed (vv) at the given displacement (x=0.03 mx = 0.03\text{ m}) using the SHM velocity formula.
v=0.80 m/sv = 0.80\text{ m/s}
The speed at displacement xx is v=ωA2x2=20×0.0520.032=20×0.0016=20×0.04=0.80 m/sv = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}.

Anahtar Kavram

Simple Harmonic Motion Velocity and Acceleration Relationships
Tahmini Süre:1m 30s
Soru 4286Soru

A body AA of mass 4.0 kg4.0\text{ kg} moving due east at a velocity of 6.0 m s16.0\text{ m s}^{-1} collides head-on with a body BB of mass 2.0 kg2.0\text{ kg} moving due west at 3.0 m s13.0\text{ m s}^{-1}. If body AA continues to move due east after the collision with a speed of 1.0 m s11.0\text{ m s}^{-1}, what is the velocity of body BB after the collision?

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Cevap: 7.0 m s17.0\text{ m s}^{-1} due East

Cevap

7.0 m s17.0\text{ m s}^{-1} due East
According to the principle of conservation of linear momentum, the total momentum before collision equals the total momentum after collision. Assigning positive to East and negative to West, total initial momentum is 4(6)+2(3)=18 kg m s14(6) + 2(-3) = 18\text{ kg m s}^{-1}. Equating this to final momentum 4(1)+2vB4(1) + 2v_B gives 2vB=142v_B = 14, resulting in vB=+7.0 m s1v_B = +7.0\text{ m s}^{-1}, which represents 7.0 m s17.0\text{ m s}^{-1} due East.

Adım Adım Çözüm

1
Define a reference direction for 1D momentum vectors
Let East be positive (+) and West be negative (-).
Linear momentum is a vector quantity, so opposite directions must have opposite algebraic signs.
2
Calculate the total initial momentum before collision (pip_i)
pi=mAuA+mBuB=(4.0 kg)(+6.0 m s1)+(2.0 kg)(3.0 m s1)=24.06.0=18.0 kg m s1p_i = m_A u_A + m_B u_B = (4.0\text{ kg})(+6.0\text{ m s}^{-1}) + (2.0\text{ kg})(-3.0\text{ m s}^{-1}) = 24.0 - 6.0 = 18.0\text{ kg m s}^{-1}.
Body B moves west, so its initial velocity is 3.0 m s1-3.0\text{ m s}^{-1}.
3
Formulate the total final momentum after collision (pfp_f)
pf=mAvA+mBvB=(4.0 kg)(+1.0 m s1)+(2.0 kg)vB=4.0+2.0vBp_f = m_A v_A + m_B v_B = (4.0\text{ kg})(+1.0\text{ m s}^{-1}) + (2.0\text{ kg})v_B = 4.0 + 2.0 v_B.
Body A moves east after collision, so its final velocity is +1.0 m s1+1.0\text{ m s}^{-1}.
4
Apply the Law of Conservation of Linear Momentum (pi=pfp_i = p_f)
18.0=4.0+2.0vB    2.0vB=14.0    vB=+7.0 m s118.0 = 4.0 + 2.0 v_B \implies 2.0 v_B = 14.0 \implies v_B = +7.0\text{ m s}^{-1}.
Since total initial momentum equals total final momentum in an isolated system.
5
Interpret the sign of the calculated velocity
Since vBv_B is positive (+7.0 m s1+7.0\text{ m s}^{-1}), body B moves at 7.0 m s17.0\text{ m s}^{-1} due East.
Positive values correspond to the defined East direction.

Anahtar Kavram

Principle of Conservation of Linear Momentum in 1D head-on collisions
Soru 4287Soru

The electric current passing through a resistor is measured as (5.0±0.1) A(5.0 \pm 0.1)\text{ A}, and the resistance is measured as (10.0±0.3) Ω(10.0 \pm 0.3)\text{ }\Omega. What is the maximum percentage error in the calculated electrical power dissipated by the resistor?

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Cevap: 7.0%7.0\%

Cevap

The maximum percentage error in the calculated power is 7.0%7.0\%.
Electrical power is calculated using the formula P=I2RP = I^2 R. In error analysis, the maximum fractional error for a quantity y=anbmy = a^n b^m is Δyy=nΔaa+mΔbb\frac{\Delta y}{y} = n\frac{\Delta a}{a} + m\frac{\Delta b}{b}. Therefore, the maximum percentage error in PP is equal to 2×(percentage error in I)+(percentage error in R)=2(2.0%)+3.0%=7.0%2 \times (\text{percentage error in } I) + (\text{percentage error in } R) = 2(2.0\%) + 3.0\% = 7.0\%.

Adım Adım Çözüm

1
Calculate the percentage error in the measured current II.
ΔII×100%=0.15.0×100%=2.0%\frac{\Delta I}{I} \times 100\% = \frac{0.1}{5.0} \times 100\% = 2.0\%
Percentage error is the ratio of absolute uncertainty to the measured value expressed as a percentage.
2
Calculate the percentage error in the measured resistance RR.
ΔRR×100%=0.310.0×100%=3.0%\frac{\Delta R}{R} \times 100\% = \frac{0.3}{10.0} \times 100\% = 3.0\%
Applying the same percentage error formula to the resistance measurement.
3
Apply the error propagation formula for power P=I2RP = I^2 R.
ΔPP×100%=2(ΔII×100%)+(ΔRR×100%)=2(2.0%)+3.0%=7.0%\frac{\Delta P}{P} \times 100\% = 2\left(\frac{\Delta I}{I} \times 100\%\right) + \left(\frac{\Delta R}{R} \times 100\%\right) = 2(2.0\%) + 3.0\% = 7.0\%
When physical quantities are raised to a power and multiplied, their fractional/percentage errors are multiplied by the exponent and summed to get the maximum error.

Anahtar Kavram

Propagation of errors in derived quantities involving powers and products

Alternatif Yöntem

Calculate absolute values of minimum and maximum possible power, then compute percentage variation relative to nominal power.
Tahmini Süre:1m 30s
Soru 4288Soru

Find the positive integer value of nn such that nP4=42×nP2^{n}P_4 = 42 \times {^{n}P_2}.

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Cevap: 9

Cevap

The positive integer value of nn is 9.
Expanding nP4^{n}P_4 as n(n1)(n2)(n3)n(n-1)(n-2)(n-3) and nP2^{n}P_2 as n(n1)n(n-1) allows dividing out n(n1)n(n-1), leading to (n2)(n3)=42(n-2)(n-3) = 42. Expanding and factoring gives n25n36=0n^2 - 5n - 36 = 0, which yields n=9n = 9 as the only valid positive integer.

Adım Adım Çözüm

1
Apply the permutation formula nPr=n!(nr)!^{n}P_r = \frac{n!}{(n-r)!}
nP4=n(n1)(n2)(n3)^{n}P_4 = n(n-1)(n-2)(n-3) and nP2=n(n1)^{n}P_2 = n(n-1)
By definition of permutations, selecting rr items from nn distinct items without replacement.
2
Substitute the expansions into the given relation
n(n1)(n2)(n3)=42n(n1)n(n-1)(n-2)(n-3) = 42 n(n-1)
Direct substitution into nP4=42×nP2^{n}P_4 = 42 \times {^{n}P_2}.
3
Simplify by dividing out common non-zero terms
(n2)(n3)=42(n-2)(n-3) = 42
Since n4n \ge 4, n(n1)0n(n-1) \neq 0 and can be safely divided from both sides.
4
Form and solve the quadratic equation
n25n+6=42    n25n36=0    (n9)(n+4)=0n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0
Expanding terms and factoring the resulting quadratic expression.
5
Determine the valid root
n=9n = 9
Permutation total items nn must satisfy nr0n \ge r \ge 0, rejecting the negative root n=4n = -4.

Anahtar Kavram

Algebraic equations involving permutations
Tahmini Süre:1m 15s
Soru 4289Soru

An electromagnetic wave has a wavelength of 1.5×107 m1.5 \times 10^{-7}\text{ m} in a vacuum. It enters an optical medium where its speed decreases to 2.0×108 m/s2.0 \times 10^8\text{ m/s}. Given that the speed of light in vacuum is c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, what is the wavelength of the wave in the medium, and to which spectral region does its frequency belong?

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Cevap: 1.0×107 m1.0 \times 10^{-7}\text{ m} and Ultraviolet region

Cevap

The wavelength in the medium is 1.0×107 m1.0 \times 10^{-7}\text{ m} and its frequency belongs to the Ultraviolet region.
When an electromagnetic wave transitions into a medium, its frequency ff remains constant at 2.0×1015 Hz2.0 \times 10^{15}\text{ Hz}, placing it strictly in the Ultraviolet spectrum. Its wavelength in the medium shrinks to λmed=vf=2.0×1082.0×1015=1.0×107 m\lambda_{med} = \frac{v}{f} = \frac{2.0 \times 10^8}{2.0 \times 10^{15}} = 1.0 \times 10^{-7}\text{ m}.

Adım Adım Çözüm

1
Calculate the frequency of the wave in vacuum using the wave equation c=fλvacc = f \lambda_{vac}.
f=3.0×108 m/s1.5×107 m=2.0×1015 Hzf = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-7}\text{ m}} = 2.0 \times 10^{15}\text{ Hz}.
Frequency is an intrinsic property determined by the source and does not change when entering a new medium.
2
Identify the electromagnetic spectrum region corresponding to the calculated frequency.
A frequency of 2.0×1015 Hz2.0 \times 10^{15}\text{ Hz} falls in the range 7.5×1014 Hz7.5 \times 10^{14}\text{ Hz} to 3.0×1016 Hz3.0 \times 10^{16}\text{ Hz}, which corresponds to the Ultraviolet region.
Spectral classifications are uniquely determined by frequency (or vacuum wavelength).
3
Determine the wavelength in the medium using λmed=vf\lambda_{med} = \frac{v}{f}.
λmed=2.0×108 m/s2.0×1015 Hz=1.0×107 m\lambda_{med} = \frac{2.0 \times 10^8\text{ m/s}}{2.0 \times 10^{15}\text{ Hz}} = 1.0 \times 10^{-7}\text{ m}.
The wavelength changes proportionally with phase velocity in a medium.

Anahtar Kavram

Invariance of electromagnetic wave frequency across media boundaries and wave equation relationship v=fλv = f \lambda.
Tahmini Süre:2m 0s
Soru 4290Soru

A stone is projected from ground level with an initial velocity of 40 m/s40\text{ m/s} at an angle of 3030^\circ to the horizontal. What is the total time of flight of the stone, in seconds? (Take g=10 m/s2g = 10\text{ m/s}^2)

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Cevap: 4

Cevap

The total time of flight of the stone is 4 s4\text{ s}.
The total time of flight TT for a projectile launched over level ground is calculated using T=2usinθgT = \frac{2 u \sin \theta}{g}. Substituting u=40 m/su = 40\text{ m/s}, θ=30\theta = 30^\circ, and g=10 m/s2g = 10\text{ m/s}^2 yields T=2×40×0.510=4 sT = \frac{2 \times 40 \times 0.5}{10} = 4\text{ s}.

Adım Adım Çözüm

1
Find the vertical component of the launch velocity
uy=20 m/su_y = 20\text{ m/s}
The vertical motion determines the time the projectile remains in the air.
2
Calculate the total time of flight
T=4 sT = 4\text{ s}
Applying T=2usinθg=2×2010=4 sT = \frac{2 u \sin \theta}{g} = \frac{2 \times 20}{10} = 4\text{ s} gives the total duration before landing back at ground level.

Anahtar Kavram

Time of Flight in Projectile Motion
Tahmini Süre:45s
Soru 4291Soru

If (x,y)(x, y) satisfies the simultaneous equations x+2y=5x + 2y = 5 and x2+y2=10x^2 + y^2 = 10, what is the positive value of xx?

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Cevap: 3

Cevap

The positive value of xx is 3.
Isolating xx in the linear equation gives x=52yx = 5 - 2y. Substituting this expression into x2+y2=10x^2 + y^2 = 10 yields (52y)2+y2=10(5 - 2y)^2 + y^2 = 10. Expanding gives 2520y+4y2+y2=10    5y220y+15=025 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0. Dividing all terms by 55 produces y24y+3=0y^2 - 4y + 3 = 0, which factors as (y1)(y3)=0(y - 1)(y - 3) = 0, so y=1y = 1 or y=3y = 3. Substituting these into x=52yx = 5 - 2y gives x=3x = 3 when y=1y = 1 and x=1x = -1 when y=3y = 3. The positive value of xx is 3.

Adım Adım Çözüm

1
Express xx from the linear equation
x=52yx = 5 - 2y
Isolating xx allows substitution into the quadratic equation.
2
Substitute into the quadratic equation
(52y)2+y2=10(5 - 2y)^2 + y^2 = 10
Eliminates variable xx to create a single-variable equation in yy.
3
Expand and simplify
5y220y+15=0    y24y+3=05y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0
Transforms the equation into standard quadratic form for easy factorization.
4
Solve for yy
y=1 or y=3y = 1 \text{ or } y = 3
Factoring (y1)(y3)=0(y - 1)(y - 3) = 0 yields the two possible values for yy.
5
Determine corresponding xx values and select the positive one
x=3x = 3 (from y=1y = 1)
Evaluating x=52yx = 5 - 2y gives x=3x = 3 and x=1x = -1; the positive result requested is 3.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution
Tahmini Süre:1m 30s
Soru 4292Soru

A satellite of mass mm moves in a circular orbit around a uniform spherical planet of radius RR. If the height of the satellite above the planet's surface is h=2Rh = 2R and the acceleration due to gravity at the planet's surface is gg, which of the following expressions represents the orbital speed of the satellite?

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Cevap: gR3\sqrt{\frac{gR}{3}}

Cevap

The orbital speed of the satellite is gR3\sqrt{\frac{gR}{3}}.
The total orbital radius from the center of the planet is r=R+h=R+2R=3Rr = R + h = R + 2R = 3R. Since surface gravity is g=GMR2g = \frac{GM}{R^2}, we have GM=gR2GM = gR^2. Substituting these into the orbital velocity expression v=GMrv = \sqrt{\frac{GM}{r}} yields v=gR23R=gR3v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}, which makes the expression gR3\sqrt{\frac{gR}{3}} correct.

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1
Determine the total orbital radius from the center of the planet.
r=R+h=R+2R=3Rr = R + h = R + 2R = 3R
Gravitational attraction and circular orbital radii are always measured from the center of mass of the primary body, not its surface.
2
Relate the gravitational constant GG and planet mass MM to surface gravity gg.
g=GMR2    GM=gR2g = \frac{GM}{R^2} \implies GM = gR^2
At the surface of a spherical planet of radius RR, the gravitational field strength is gg.
3
Substitute r=3Rr = 3R and GM=gR2GM = gR^2 into the circular orbital velocity equation v=GMrv = \sqrt{\frac{GM}{r}}.
v=gR23R=gR3v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}
Equating centripetal force to gravitational force mv2r=GMmr2\frac{m v^2}{r} = \frac{G M m}{r^2} yields v=GMrv = \sqrt{\frac{GM}{r}}.

Anahtar Kavram

Orbital velocity of a satellite in terms of surface gravitational acceleration and orbital radius
Soru 4293Soru

A motorist traveling along a straight highway at a constant speed of 30 m/s30\text{ m/s} observes a road hazard ahead. The driver experiences a reaction delay of 0.5 s0.5\text{ s} before applying the brakes, after which the vehicle decelerates uniformly at a rate of 5 m/s25\text{ m/s}^2. What is the total distance traveled by the vehicle from the instant the hazard is observed until the vehicle comes to a complete stop?

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Cevap: 105 m105\text{ m}

Cevap

The total distance traveled by the vehicle before coming to a stop is 105 m105\text{ m}.
The motion consists of two distinct stages: a constant-speed phase during the 0.5 s0.5\text{ s} reaction time (15 m15\text{ m}) and a uniformly decelerating phase until rest (90 m90\text{ m}). Summing both components yields 105 m105\text{ m}.

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1
Calculate the reaction distance (s1s_1) covered during the driver's reaction delay.
s1=u×tr=30 m/s×0.5 s=15 ms_1 = u \times t_r = 30\text{ m/s} \times 0.5\text{ s} = 15\text{ m}
Before the brakes are applied, the vehicle continues moving at its initial constant speed of 30 m/s30\text{ m/s}.
2
Calculate the braking distance (s2s_2) using the third equation of motion.
v2=u2+2as2    02=302+2(5)s2    10s2=900    s2=90 mv^2 = u^2 + 2as_2 \implies 0^2 = 30^2 + 2(-5)s_2 \implies 10s_2 = 900 \implies s_2 = 90\text{ m}
The vehicle decelerates from 30 m/s30\text{ m/s} to a final velocity of 0 m/s0\text{ m/s} at a uniform deceleration rate of a=5 m/s2a = -5\text{ m/s}^2.
3
Sum the reaction distance and the braking distance to find the total stopping distance.
stotal=s1+s2=15 m+90 m=105 ms_{\text{total}} = s_1 + s_2 = 15\text{ m} + 90\text{ m} = 105\text{ m}
The total stopping distance is the sum of the distance covered before braking begins and the distance covered while braking.

Anahtar Kavram

Two-phase linear motion combining constant velocity reaction distance and uniform deceleration braking distance.
Soru 4294Soru

What is the indefinite integral ((3x+1)26sin(3x))dx\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx?

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Cevap: 3x3+3x2+x+2cos(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C

Cevap

3x3+3x2+x+2cos(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C
Expanding (3x+1)2(3x+1)^2 yields 9x2+6x+19x^2 + 6x + 1. Integrating 9x2+6x+19x^2 + 6x + 1 gives 3x3+3x2+x3x^3 + 3x^2 + x. Integrating 6sin(3x)-6\sin(3x) gives +2cos(3x)+2\cos(3x). Combining these results with the mandatory constant of integration +C+ C gives 3x3+3x2+x+2cos(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C.

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1
Expand the squared polynomial binomial term
(3x+1)2=9x2+6x+1(3x + 1)^2 = 9x^2 + 6x + 1
Expanding the expression allows for simple term-by-term integration using the power rule.
2
Integrate the polynomial component term by term
(9x2+6x+1)dx=3x3+3x2+x\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x
Apply the power rule xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each polynomial term.
3
Integrate the trigonometric component
6sin(3x)dx=6(13cos(3x))=+2cos(3x)\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)
The integral of sin(kx)\sin(kx) is 1kcos(kx)-\frac{1}{k}\cos(kx). Multiplying by 6-6 yields +2cos(3x)+2\cos(3x).
4
Combine terms and add the constant of integration
3x3+3x2+x+2cos(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C
Indefinite integrals must include an arbitrary constant +C+ C to account for all antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions
Soru 4295Soru

A man standing at a distance in front of a tall vertical cliff claps his hands and hears the echo after 1.2 s1.2\text{ s}. If the speed of sound in air is 340 m/s340\text{ m/s}, what is the distance between the man and the cliff in meters?

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Cevap: 204

Cevap

The distance between the man and the cliff is 204 m204\text{ m}.
An echo is formed when sound travels to an obstacle and reflects back. The time taken for the sound to travel to the cliff and back is 1.2 s1.2\text{ s}. The total distance covered by sound is v×t=340 m/s×1.2 s=408 mv \times t = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}. Since this distance covers two equal trips (to the cliff and back), the distance to the cliff is 408 m/2=204 m408\text{ m} / 2 = 204\text{ m}.

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1
State the relationship between sound speed, total echo time, and distance.
Total distance traveled by the sound is twice the distance to the cliff: 2d=v×t2d = v \times t.
An echo involves sound traveling from the source to the reflecting barrier and back.
2
Substitute the given values into the equation.
2d=340 m/s×1.2 s=408 m2d = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}.
To find the total distance traversed by the sound wave.
3
Solve for the distance dd.
d=408 m2=204 md = \frac{408\text{ m}}{2} = 204\text{ m}.
The one-way distance to the cliff is half of the total distance traveled by the echo.

Anahtar Kavram

Calculation of echo distance using d=vt2d = \frac{v t}{2}
Soru 4296Soru

If the expression 353+5\frac{3 - \sqrt{5}}{3 + \sqrt{5}} is simplified and written in the form a+b5a + b\sqrt{5}, where aa and bb are rational numbers, what is the numerical value of a+ba + b?

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Cevap: 2

Cevap

The numerical value of a+ba + b is 22.
To express 353+5\frac{3 - \sqrt{5}}{3 + \sqrt{5}} in the standard form a+b5a + b\sqrt{5}, multiply both the numerator and denominator by the conjugate of the denominator, which is (35)(3 - \sqrt{5}). The numerator expands to (35)2=965+5=1465(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}. The denominator becomes 32(5)2=95=43^2 - (\sqrt{5})^2 = 9 - 5 = 4. Dividing gives 144645=72325\frac{14}{4} - \frac{6}{4}\sqrt{5} = \frac{7}{2} - \frac{3}{2}\sqrt{5}. Hence, a=72a = \frac{7}{2} and b=32b = -\frac{3}{2}, making a+b=7232=42=2a + b = \frac{7}{2} - \frac{3}{2} = \frac{4}{2} = 2.

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1
Multiply numerator and denominator by the conjugate of the denominator
\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}
To eliminate the surd from the denominator.
2
Expand both the numerator and the denominator
14654\frac{14 - 6\sqrt{5}}{4}
Using (xy)2=x22xy+y2(x-y)^2 = x^2 - 2xy + y^2 for the numerator and difference of two squares (x+y)(xy)=x2y2(x+y)(x-y) = x^2 - y^2 for the denominator.
3
Separate into rational component and radical coefficient
72325\frac{7}{2} - \frac{3}{2}\sqrt{5}
Simplifying fractions by dividing numerator and denominator by their greatest common divisor.
4
Calculate the sum a+ba + b
7232=2\frac{7}{2} - \frac{3}{2} = 2
Comparing 72325\frac{7}{2} - \frac{3}{2}\sqrt{5} with a+b5a + b\sqrt{5} yields a=72a = \frac{7}{2} and b=32b = -\frac{3}{2}.

Anahtar Kavram

Rationalization of Binomial Denominators using Conjugates
Tahmini Süre:1m 30s
Soru 4297Soru

A metallic sphere weighs 5.0 N5.0\text{ N} in air. When completely immersed in water, its apparent weight is 3.0 N3.0\text{ N}. When completely immersed in an unknown liquid XX, its apparent weight is 3.4 N3.4\text{ N}. What is the density of liquid XX in kg/m3\text{kg/m}^3? (Take the density of water as 1000 kg/m31000\text{ kg/m}^3 and acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2).

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Cevap: 800

Cevap

The density of liquid XX is 800 kg/m3800\text{ kg/m}^3.
The upthrust in water (2.0 N2.0\text{ N}) gives the volume of the sphere as 2.0×104 m32.0 \times 10^{-4}\text{ m}^3. Using the upthrust in liquid X (1.6 N1.6\text{ N}), the density of liquid X is calculated as ρX=1.6(2.0×104)(10)=800 kg/m3\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3.

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1
Calculate upthrust in water
Uw=5.0 N3.0 N=2.0 NU_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}
Upthrust equals the loss in weight of the submerged body in water.
2
Determine the volume of the metallic sphere
V=Uwρwg=2.01000×10=2.0×104 m3V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3
According to Archimedes' principle, upthrust in water equals the weight of displaced water.
3
Calculate upthrust in liquid X
UX=5.0 N3.4 N=1.6 NU_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}
Loss of weight in liquid X gives the upthrust exerted by liquid X.
4
Calculate the density of liquid X
ρX=UXVg=1.6(2.0×104)×10=800 kg/m3\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3
Rearranging UX=ρXVgU_X = \rho_X V g allows solving for the unknown fluid density.

Anahtar Kavram

Archimedes' Principle and Apparent Weight
Tahmini Süre:2m 0s
Soru 4298Soru

An electric heater rated at 50 W50\text{ W} is used to heat a solid block of mass 1.5 kg1.5\text{ kg} for 4 minutes4\text{ minutes}. During this period, the temperature of the block increases from 30C30^\circ\text{C} to 70C70^\circ\text{C}. If 20%20\% of the heat energy supplied by the heater is lost to the surroundings, what is the heat capacity of the block?

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Cevap: 240 J K1240\text{ J K}^{-1}

Cevap

240 J K1240\text{ J K}^{-1}
The correct answer is 240 J K1240\text{ J K}^{-1}. The heater delivers 12000 J12{}000\text{ J} of energy in 4 minutes4\text{ minutes}. Accounting for 20%20\% environmental heat loss leaves 9600 J9{}600\text{ J} absorbed by the block. Dividing this energy by the 40 K40\text{ K} temperature rise gives a total heat capacity of 240 J K1240\text{ J K}^{-1}.

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1
Calculate total electrical energy supplied by heater
Qsupplied=P×t=50 W×(4×60 s)=12000 JQ_{\text{supplied}} = P \times t = 50\text{ W} \times (4 \times 60\text{ s}) = 12{}000\text{ J}
Electrical work converted to heat is given by power multiplied by time in seconds.
2
Determine useful heat absorbed by block after accounting for energy loss
Quseful=(10.20)×12000 J=0.80×12000 J=9600 JQ_{\text{useful}} = (1 - 0.20) \times 12{}000\text{ J} = 0.80 \times 12{}000\text{ J} = 9{}600\text{ J}
Since 20%20\% of supplied energy is lost, 80%80\% is retained to raise the temperature of the block.
3
Calculate temperature rise
ΔT=70C30C=40 K\Delta T = 70^\circ\text{C} - 30^\circ\text{C} = 40\text{ K}
Temperature difference is the final temperature minus the initial temperature.
4
Calculate heat capacity of the block
C=QusefulΔT=9600 J40 K=240 J K1C = \frac{Q_{\text{useful}}}{\Delta T} = \frac{9{}600\text{ J}}{40\text{ K}} = 240\text{ J K}^{-1}
Heat capacity CC is defined as total heat absorbed per unit temperature change (C=QΔTC = \frac{Q}{\Delta T}).

Anahtar Kavram

Heat Capacity (C=QΔTC = \frac{Q}{\Delta T}) represents total thermal capacity of a body, whereas Specific Heat Capacity (c=QmΔTc = \frac{Q}{m\Delta T}) is heat capacity per unit mass.
Soru 4299Soru

A student stands between two tall, parallel vertical walls and claps her hands once. She hears the first echo after 1.0 s1.0\text{ s} and the second echo after 1.5 s1.5\text{ s}. Taking the speed of sound in air to be 340 m s1340\text{ m s}^{-1}, what is the distance between the two walls?

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Cevap: 425 m425\text{ m}

Cevap

The distance between the two walls is 425 m425\text{ m}.
Sound travels from the student to each wall and reflects back. The distance to the first wall is 340×1.02=170 m\frac{340 \times 1.0}{2} = 170\text{ m} and to the second wall is 340×1.52=255 m\frac{340 \times 1.5}{2} = 255\text{ m}. Adding both distances gives the total separation between the walls as 425 m425\text{ m}.

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1
Calculate the distance from the student to the closer wall (d1d_1)
d1=v×t12=340 m s1×1.0 s2=170 md_1 = \frac{v \times t_1}{2} = \frac{340 \text{ m s}^{-1} \times 1.0 \text{ s}}{2} = 170\text{ m}
Sound travels to the wall and reflects back, so the time given corresponds to twice the distance.
2
Calculate the distance from the student to the further wall (d2d_2)
d2=v×t22=340 m s1×1.5 s2=255 md_2 = \frac{v \times t_2}{2} = \frac{340 \text{ m s}^{-1} \times 1.5 \text{ s}}{2} = 255\text{ m}
The sound for the second echo travels to the second wall and back.
3
Determine the total distance between the two parallel walls
D=d1+d2=170 m+255 m=425 mD = d_1 + d_2 = 170\text{ m} + 255\text{ m} = 425\text{ m}
Since the student is positioned between the two walls, the separation of the walls is the sum of both individual distances.

Anahtar Kavram

Echo distance relation 2d=vt2d = v t for sound reflection from barriers.
Tahmini Süre:1m 30s
Soru 4300Soru

A metal rod has a linear expansivity of 1.5×105 K11.5 \times 10^{-5}\text{ K}^{-1}. What is the volume expansivity of a sphere made from the same metal?

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Cevap: 4.5×105 K14.5 \times 10^{-5}\text{ K}^{-1}

Cevap

The volume expansivity of the sphere is 4.5×105 K14.5 \times 10^{-5}\text{ K}^{-1}.
The volume expansivity γ\gamma of a uniform solid object is related to its linear expansivity α\alpha by γ=3α\gamma = 3\alpha. Multiplying 1.5×105 K11.5 \times 10^{-5}\text{ K}^{-1} by 33 gives 4.5×105 K14.5 \times 10^{-5}\text{ K}^{-1}.

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1
Identify the mathematical relationship between linear expansivity (α\alpha) and volume (cubical) expansivity (γ\gamma).
γ=3α\gamma = 3\alpha
For an isotropic solid, volume expansion occurs equally in three dimensions, making the volume coefficient three times the linear coefficient.
2
Substitute the given value of linear expansivity into the formula and calculate.
γ=3×(1.5×105 K1)=4.5×105 K1\gamma = 3 \times (1.5 \times 10^{-5}\text{ K}^{-1}) = 4.5 \times 10^{-5}\text{ K}^{-1}
Obtain the numeric value of volume expansivity.

Anahtar Kavram

Relationship between linear and volume expansivity of solids
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