高三選修物理 請問第三小題的10是從哪裡來呢？ 麻煩大家ㄌ！謝謝回答🥹🫶

| 進階題 *為多選題 12. 有一線圈在均勻磁場中轉動時,通過線圈面的磁通量史 與時間成正弦關係,如右圖所示。 試回答下列問題: (1)此線圈轉動的頻率為多少rad/s? 元 (A)z (B)- (C) (D) (E) 8 16 2 4 (2)瞬時應電動勢與時間的關係圖為下列何者? (A) ε W (B) E (C)e $ (Wb) 15 10 5 ww -5 -10 15 0 2 4 6 8 A 10 12 14 時間(s) (D) & (E) & m.. BA 3)此線圈上瞬時應電動勢的最大值為多少 V? A45% 10x = =570 E (4)在下列的哪一個瞬間,此線圈上應電動勢的值最大?, (A)第0.4秒(B)第0.8秒(C)第1.2秒(D)第1.6秒(E)第2.0秒 (A)第6.4 (5)第1.0秒時,線圈上的瞬時應電動勢為多少? 0 A ☑ (6)在2.0秒至3.0 秒期間,線圈上平均應電動勢量值為多少V? (A) 0 (B) 2 (C) 4 (D) 8 (E) 10 3-23/0

請問這題怎麼算呢？謝謝～

3. 如圖某球體質量M,半徑R,求將其鏤空出一直徑為R的小球後,將一質量為m 之質點放在C、D兩點時,所受之重力為何? Ans: C點:GMm/2R²←← D點:7GMm/36R2- C= D R

求解🙏🏻

挑戰一:某連隊等速直線前進,部隊全長 600m,傳令兵自隊尾等速跑到隊首遞交報告給連長後, 立刻以原速率跑回隊尾,此時整個連隊恰好前進了800m,則全程傳令兵移動的路徑長為?/boom # A popular web video shows a jet airplane, a car, and a motorcycle racing from rest along a runway. Initially the motorcycle takes the lead, but then the jet takes the lead, and finally the car blows past the motorcycle. Here let's focus on the car and motorcycle and assign some reasonable values to the motion. The motorcycle first takes the lead because its (constant) acceleration am 8.40 m/s² is greater than the car's (constant) acceleration ac 5.60 m/s², but it soon loses to the car because it reaches its greatest 58.8 m/s before the car reaches its greatest speed 106 m/s. How long does speed the car take to reach the motorcycle? 16.6

這題為什麼F=mg?如果是水平面的話不是會和正向力抵銷嗎

三 9. 有一彈簧長,將一端固定而另一端鉛直懸掛一質量為 m 2000 L Usefzalet 的物體時會伸長達到平衡。今將彈簧一端固定,另一端 18.08 4 繫相同物體在光滑水平桌面上作等速圓周運動,如圖所示。 (重力加速度量值為g) (1) 彈簧的彈性常數為何?(5分) (1)F=kx (E) L 4 mg = kx 4 (共):一單 (2) 若物體作等速圓周運動時,彈簧的長度為L,則物體旋轉的速率為何?(5分) 疫局2 X041141 m01 3 答: MX M M4E-1 Evoica) (1) 2018) (A) OF XL 2 飯、水晶消化、庭 本商木( ) 二 4mg 二、多重選擇題:(10分,將品變速高半數大類的戰部爾麻木( ) 【VEN D) (8) I(A)

C

Which statement is not correct according to the "theory of special relativity"? (A) Whether two events at different locations are simultaneous depends on the frame; (B) The relative velocity of two objects never precede the speed of light; (C) The "effect" may appear to occur earlier than its "cause" as observed in some frames; (D) All; (E) None.

請問第二張照片紅線那邊 為什麼時間是用兩次撞擊的時間差算 不是應該撞擊瞬間的時間（碰到A面的時間）

45 (30 為 泡草 J 400 速 間沒 在一 總能 有位 可視 9 日世 況 T XCX PM 時 -V = 174 高中學習講義 選修物理 II 學習概念2 碰撞與氣體壓力的微觀論點推導(配合課本p.206) 1. 密閉容器內的氣體壓力: (1) 條件:氣體是由數目極龐大的分子所組成。(若分子數目太少時,分子行為可能偏離 平均值) 相等。 且分子的運動是「隨機」的,任一時段內向各方向運動的平均分子數目均 (2) 壓力來源: ⓊP:氣體分子的重量所造成。P:氣體分子不斷撞擊器壁所造成。 2 由於碰撞力遠大於分子重量故 P, > P,所以 P, 可忽略不計。 2. 平均壓力理論推導: (1) 說明:設邊長L、體積V(=L)的正立方體密閉容器內,有N 個相同的氣體分子,每個分子的質量均為m,如右圖。 (2)推導過程: 設將氣體分子逐一編號,其中編號第i個的氣體分子以 ^ v=vxi + urj + v.k的速度,入射 A, 面,如圖(一)因分子與器壁的碰撞為完全 彈性碰撞,則該分子與 A, 面碰撞後, 的速度分量 vi因與 4. 面垂直,故在碰撞 後變成 - va。而在y與z軸的速度分量 與vÊ不變,如圖(二)所示。 -Ux- U₁ A₂ Ax Uy U2OUx x ▲圖(一):第i顆氣體分子以速度朝4. ▲圖(二):分子撞擊器壁,其速度變化的情形。 面入射。 ② 則該分子的動量變化為Apx 2mus,即此分子在此次碰撞後,受到 4 面的衝量 == 為: J=Ap=-2musi (3) 當分子從 4. 面反彈後,有可能撞擊其他器壁面,因彈性碰撞,故其平行於x軸的速 度分量 vi速率不變,因此來回一次撞擊 4. 面,所經歷路程為 2L,共歷時 A t = 2L 。 Vix 故此第i個分子,施於面的平均力為: 015 Uz

第三小題錯在哪裡

3.)* True/false: Find the incorrect conclusions from the following sentences and correct them. (12 points) b H2 •T2 H An ideal gas changes its states isothermally through a chemical reaction. AH is zero because AH is equal to Cp x AT. H TI ® The pressure change does not contribute to the energy state (U or H) of ideal gases. P * Figure 1 is correct for illustrating the enthalpy state of real gases. Fig. 1 AT1 ATL+1a na:daalraa ahonnan kantot fo. U +- IT 1.1 D1

第一題 怎麼想

1.) True/false, multiple choice or short answer. “*” denotes more than one correct answers. (33 points) (a) Multiple Choice: I will have the largest energy transfer as expansion work from an ideal gas system 1o the surroundings if I use a process that is (3 points) (1) Isothermal and quick ;(2) adiabatic and quick ;( 3) isothermal and slow; (4) adiabatic and slow

21題 教教我QAQ

Section 11.3 Angular Momentum J-s behind rmv a he ur 265 = ? a ter 30. Example 11.1: A 58.1 Iw COMP string and whirled at 18. Express the units of angular momentum (a) using only the funda- momentum of magr mental units kilogram, meter, and second; (b) in a form involving the circular path. Fir newtons; (c) in a form involving joules. marw horizontal and (b) the 19. Use data from Appendix E to make an order-of-magnitude esti- 31. Example 11.2. A stai mate for the angular momentum of our Solar System about the Max 27 the end of its lifetime galactic center. mu = 17.293 of radius 4.96 X 10' 20. A gymnast of rotational inertia 63 kg•m? is tumbling head over heels dwarf of radius 4.21 with angular momentum 460 kg•m?/s. What's her angular speed? 17.99 acted on the core, fine A 660-g hoop 95 cm in diameter is rotating at 170 rpm about its 32. Example 11.2: Astro 22. A 1.3-th-diameter golf ball has mass 45 g and is spinning at central axis. What's its angular momentum? L: 1W=0-66*(0-95) x 140x277.10 km and determin core that collapsed to 3000 rpm. Treating the golf ball as a uniform solid sphere, what's ing with a period of 4 its angular momentum? 18-3 33. Example 11.2. The s 10-598 rpm With her arms outstre Section 11.4 Conservation of Angular Momentum tational inertia is 3.5 23. A potter's wheel with rotational inertia 6.20 kg•m? is spinning (Fig. 11.6b), her rot freely at 20.0 rpm. The potter drops a 2.50-kg lump of clay onto her final spin rate? L = 6 2 x 2-1 = 12-99~13 - 2x2 22 13= 60.48)" x 2.5 W W = 2 . X z W = 20x2T 60 ²2.1 ~ 1 12.574

第21題 到底要怎麼寫ಥ_ಥ

Section 11.3 Angular Momentum J-s behind rmv a he ur 265 = ? a ter 30. Example 11.1: A 58.1 Iw COMP string and whirled at 18. Express the units of angular momentum (a) using only the funda- momentum of magr mental units kilogram, meter, and second; (b) in a form involving the circular path. Fir newtons; (c) in a form involving joules. marw horizontal and (b) the 19. Use data from Appendix E to make an order-of-magnitude esti- 31. Example 11.2. A stai mate for the angular momentum of our Solar System about the Max 27 the end of its lifetime galactic center. mu = 17.293 of radius 4.96 X 10' 20. A gymnast of rotational inertia 63 kg•m? is tumbling head over heels dwarf of radius 4.21 with angular momentum 460 kg•m?/s. What's her angular speed? 17.99 acted on the core, fine A 660-g hoop 95 cm in diameter is rotating at 170 rpm about its 32. Example 11.2: Astro 22. A 1.3-th-diameter golf ball has mass 45 g and is spinning at central axis. What's its angular momentum? L: 1W=0-66*(0-95) x 140x277.10 km and determin core that collapsed to 3000 rpm. Treating the golf ball as a uniform solid sphere, what's ing with a period of 4 its angular momentum? 18-3 33. Example 11.2. The s 10-598 rpm With her arms outstre Section 11.4 Conservation of Angular Momentum tational inertia is 3.5 23. A potter's wheel with rotational inertia 6.20 kg•m? is spinning (Fig. 11.6b), her rot freely at 20.0 rpm. The potter drops a 2.50-kg lump of clay onto her final spin rate? L = 6 2 x 2-1 = 12-99~13 - 2x2 22 13= 60.48)" x 2.5 W W = 2 . X z W = 20x2T 60 ²2.1 ~ 1 12.574