物理評核 2026-08-06

conduction / radiation — The student correctly stated radiation as a heat transfer method.

State 'Will not'. — The student correctly stated 'No', which is equivalent to 'Will not'.

Substitution: E = (0.40 x 10^-3)(2.26 x 10^6) — Correct substitution of mass and latent heat into the equation.

Answer: 904 J — Correct answer with unit.

Substitution: Ek = 1/2 (6.63 x 10^-26)(500)^2 — Correct substitution into Ek formula.

Answer: 8.29 x 10^-21 J (Accept: 8.29 ~ 8.30 x 10^-21 J) — Correct final answer with correct unit.

State that temperature remains at T0. — Correctly stated that the temperature remains at T0.

Substitution of values into work-energy equation (Fs = 1/2mv^2) or kinematic equations with F=ma. — Correct substitution into F=ma and kinematic equations.

Final answer v = 1.5 m s^-1. — Correct final speed.

[1/2] Correct identification and direction of at least two forces (Normal reaction upwards, Weight downwards, Friction to the left). — 學生在方塊 P 上畫出三個有箭嘴的力向量：由方塊頂部向上之垂直箭嘴標註「Normal force」、由方塊底部向下之垂直箭嘴標註「weight」，以及由方塊右側水平向左之箭嘴標註「friction」。因此至少兩個力的名稱及方向均明確正確。

[2/2] Correct identification and direction of all three forces with appropriate labels. — 學生的三個答案箭嘴均由方塊 P 邊緣出發：正常力箭嘴垂直向上、重量箭嘴垂直向下、摩擦力箭嘴水平向左；三者分別清楚標有「Normal force」、「weight」和「friction」。這與 MS 新加的三個力箭嘴及標註一致。

Substitution into vertical displacement formula s = 1/2 gt^2 using g = 9.81 and t = 0.35. — Correct substitution into vertical displacement formula using g = 9.81 and t = 0.35.

Final answer for height in range 0.60 m to 0.613 m. — Final answer 0.601 m is within the accepted range of 0.60 m to 0.613 m.

Substitution into centripetal force formula F = mω^2r. — Correct substitution into centripetal force formula F = m * r * w^2.

Final answer 0.36 N. — Correct final magnitude of 0.36 N.

State that they have the same angular speed. — Correctly states that P and Q have the same angular speed.

State that the magnitude of centripetal acceleration is different or smaller for Q. — Correctly states that P and Q have different magnitudes of centripetal acceleration.

Explain that this is because Q has a different or smaller radius (rQ < rP). — Correctly explains that the difference is due to different radius.

Identify point R as the particle moving with the maximum speed. — The student correctly identified region R.

Correct substitution or calculation to find the diffraction angle θ (e.g., using tan θ = (x/2)/L). — The student correctly used tan θ = (x/2)/L to find the diffraction angle θ = 24.7°.

Correct substitution into the grating equation d sin θ = mλ. — The student correctly substituted values into the grating equation d sin θ = mλ.

Final answer for d is approximately 1.56 μm (Accept: 1.56 μm ~ 1.6 μm). — The student obtained the correct final answer for d as 1.56×10⁻⁶ m.

State that the distance x will decrease. — The student correctly stated that x will decrease.

Explain that as wavelength λ decreases, the diffraction angle θ (and sin θ) decreases. — The student correctly explained using the grating formula that a decrease in wavelength λ leads to a decrease in the separation x.

Calculation of speed ratio (approx 8.90 x 10^5). — Correctly calculated speed ratio as 8.90 x 10^5.

Substitution into distance = speed x time (337 x 3.0). — Correct substitution into distance = speed x time.

Final answer 1011 m (Accept 1010 m ~ 1011 m). — Correct final answer of 1011 m with proper unit.

Substitution into Snell's Law: 1.33 = sin r / sin 1.5°. — Correct substitution into Snell's law equivalent to 1.33 = sin r / sin 1.5°.

Final answer r ≈ 2.00°. — Correct final angle of refraction of 2.00°.

Method using trigonometry (tan ratios) or the small angle approximation (real depth / apparent depth = n) to relate h and h1. — Correct method using small angle approximation ratio (apparent depth = real depth / n).

State that the charge remains unchanged. — Correctly states that the amount of charges on the rod does not decrease.

Explain that charging is through induction without contact or transfer of charges. — Correctly explains that process involves no contact and no transfer of charges.

Correct procedure to identify polarity: repulsion indicates positive charge; attraction indicates negative charge (1M for the correct procedure). — Correct procedure explaining that repulsion identifies same polarity and attraction identifies opposite polarity.

Measure/mark a specific release angle using a protractor. — The student mentions setting the ball at a specific angle using the protractor.

Release the ball from rest with the string taut. — The student describes releasing the ball from rest with the string kept taut (maximum tension).

Calculate velocity of E using momentum conservation: vE = 1.0 m s^-1. — The student correctly applies momentum conservation to find the velocity vE = 1.0 m s^-1.

State that the collision is not perfectly elastic. — The student correctly states that the collisions are not perfectly elastic.

Conclude that resistance increases according to R = ρl/A. — The student correctly concluded that resistance increases according to R = ρl/A.

Correct expression for Vin: Vin = I1(R1 + R2). — Correct relation formula between Vin, I1, R1, and R2.

Substitution of values into the bridge circuit ratio formula for both 'before' and 'after' states. — Correct substitution of values into the bridge circuit formula for both unstretched and stretched states.

Calculate the percentage increase as approximately 3.23% (Accept ±(3.2~3.23)%). — Correctly calculated the percentage change as 3.23%, which falls within the acceptable range.

Identify crack ①. — Correctly identified crack ①.

Identify direction along AB. — Correctly identified direction AB.

Correct field pattern (lines passing through the center and looping around). — 學生在 Figure 12.1 的線圈周圍加畫了多條連續的場線：兩側有由線圈上方繞到下方的大型圓滑迴圈，左右各有較小的內迴圈；另有近乎豎直的中間場線穿過線圈的中央區域，再向線圈兩側／外圍回繞。這些新增線條不是題目原有的橢圓線圈，並表達了場線穿過線圈中心及在外側回繞的線圈磁場閉合迴路圖樣。

Correct direction of field lines (pointing downwards through the center of the coil). Note: Also accept field lines pointing upwards for observers with different perception of current direction. — 學生在穿過線圈中央的豎直場線上畫有清楚箭嘴；在線圈中心下方的兩條近豎直線上，箭頭尖端均向下，表示磁場在線圈中心沿向下方向通過。這與 MS 圖中新加的中央場線箭頭向下相符；外側回繞線上的箭頭方向亦與此閉合方向一致。

Correct frequency (frequency doubled). — Inside the printed CRO rectangle, the student has added a blue sinusoidal trace with about four complete oscillations across the same horizontal time frame. The original printed black trace has about two complete oscillations across that frame, so the student's added trace explicitly shows doubled frequency.

The a.c. flowing in transmitter coil T generates a changing magnetic field. — Correctly stated that alternating current generates a time-varying magnetic field.

By electromagnetic induction, an induced e.m.f. is produced in receiver coil R to oppose the changing magnetic flux experienced by it. — Correctly stated that an induced e.m.f. is produced in coil R to oppose the changing magnetic flux.

Substitution into k = ln 2 / t_1/2 with conversion of days to seconds. — Correct substitution into formula with conversion of days to seconds.

Final answer: 2.10 * 10^-6 (s^-1). — Correct final numerical value and unit.

Substitution into A = kN. — Correct substitution into equation A = kN.

Final answer: 2.29 * 10^7 (Accept: 2.28 ~ 2.3 * 10^7). — Final answer 2.29 x 10^7 is within the accepted range (2.28 ~ 2.3 x 10^7).

Strong ionizing power of alpha particles emitted might affect organs/cells nearby. — Correctly identified the high ionizing power of alpha particles.