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{"id": "CAP_physics_dev-112-9", "question": "舞臺劇演出時,通常會讓周遭的環境昏暗,再用聚光燈來照射演員,讓觀眾能看見演員的表演。有關觀眾能看見演員表演的敘述,下列何者最合理?", "A": "聚光燈發出的光線照射在演員上,演員吸收這些光線,因此觀眾能看見演員", "B": "觀眾眼睛發出的光線照射在演員上,演員折射這些光線,因此觀眾能看見演員", "C": "聚光燈發出的光線照射在演員上,演員折射這些光線,因此觀眾能看見演員", "D": "聚光燈發出的光線照射在演員上,演員反射這些光線,因此觀眾能看見演員", "E": null, "F": null, "answer": "D", "explanation": "能看見物體,是因光源將光線照在物體之上,然後反射的光進入觀眾的眼睛,因此觀眾才能看見演員", "metadata": {"timestamp": "2023-10-23 18:52:18.204379", "source": "CAP_physics-112", "explanation_source": "https://public.ehanlin.com.tw/pre-exam/cap/112%E6%9C%83%E8%80%83%E8%87%AA%E7%84%B6%E8%A7%A3%E6%9E%90.pdf"}, "human_evaluation": {"quality": "", "comments": ""}} |
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{"id": "CAP_physics_dev-111-2", "question": "成熟的蓮霧會自然從樹上掉落到地面,蓮霧在掉落的過程中,其速率逐漸增加。上述現象是下列何種能量減少而轉換成其他形式的能量所造成的?", "A": "重力位能", "B": "動能", "C": "熱能", "D": "彈力(性)位能", "E": null, "F": null, "answer": "A", "explanation": "物體由高處落下,重力位能減少,轉變為動能和熱能。 ", "metadata": {"timestamp": "2023-10-23 18:52:18.204471", "source": "CAP_physics-111", "explanation_source": "https://app.hle.com.tw/111Exam/%E7%BF%B0%E6%9E%97%E5%87%BA%E7%89%88-111%E5%B9%B4%E5%9C%8B%E4%B8%AD%E6%9C%83%E8%80%83%E8%A7%A3%E6%9E%90%E5%8D%B7(%E8%87%AA%E7%84%B6%E7%A7%91).pdf"}, "human_evaluation": {"quality": "", "comments": ""}} |
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{"id": "CAP_physics_dev-108-34", "question": "某處化學藥品倉庫發生爆炸,網路上出現很多目擊者拍攝的影片,其中一位目擊者當時拍攝的位置距離爆炸位置約 1.5 km,則有關此目擊者所拍攝的影片,下列描述何者最合理?", "A": "影片中聽到爆炸聲後約經過 0.04~0.05 秒才看到此爆炸的爆炸火光", "B": "影片中看到爆炸火光後約經過 0.04~0.05 秒才聽到此爆炸的爆炸聲", "C": "影片中看到爆炸火光後約經過 4~5 秒才聽到此爆炸的爆炸聲", "D": "影片中聽到爆炸聲後約經過 4~5 秒才看到此爆炸的爆炸火光", "E": null, "F": null, "answer": "C", "explanation": "本題主要是比較光速與聲速的快慢差異。\n光速為光傳遞的速度約為 $3 \\times 10^8$ m/s,\n聲音在空氣中傳播的速率約為343 m/s,\n當發生爆炸時,爆炸的火光與爆炸聲是同時發生,\n光速傳遞的速度太快,遠大於聲速343 m/s,所以在遠處是先看到爆炸火光,再聽到爆炸聲。\n因為光傳遞的時間極短,所以只需計算聲音在空氣中傳播的時間,\nS:距離=1.5 km=1500 m\nV:聲速= 約343 m / s\nt:時間 t = S/V = 1500/343 = 約 4.373 (s)。\n所以答案選C。", "metadata": {"timestamp": "2023-10-23 18:52:18.204701", "source": "CAP_physics-108", "explanation_source": "https://twoliunature.pixnet.net/blog/post/9068867-108%E5%9C%8B%E4%B8%AD%E6%9C%83%E8%80%83%E8%87%AA%E7%84%B6%E7%A7%91%E7%AC%AC34%E9%A1%8C"}, "human_evaluation": {"quality": "", "comments": ""}} |
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{"id": "CAP_physics_dev-106-45", "question": "在某一溫度下,有一杯重量百分濃度 40\\%的檸檬酸水溶液 150g,再加入檸檬酸 65g 攪拌過濾,將濾紙烘乾並秤重後,發現有 5g 檸檬酸未溶解。若過程中溶液溫度均未改變,則在此溫度時檸檬酸的溶解度最接近下列何者?", "A": "91g/100g水", "B": "80g/100g水", "C": "133g/100g水", "D": "45g/100g水", "E": null, "F": null, "answer": "C", "explanation": "本題主要是問溶解度的計算。\n檸檬酸水溶液重量百分濃度 = 40\\% , 檸檬酸水溶液重 = 150 g\n溶質重(檸檬酸)= 檸檬酸水溶液重 × 重量百分濃度(\\%)= 150 × 40% = 60 g\n溶劑重(水) = 150 – 60 = 90 g\n再加入溶質(檸檬酸) 65 g,但有 5g 溶質(檸檬酸)未溶解,\n所以溶質重(檸檬酸) = 60 + 65 - 5 = 120 g。\n所以90g的水最多溶解檸檬酸120 g。\n設100公克的水,最多溶解檸檬酸 x g,120 / 90 = x / 100\nx= 約133.33 g,故(C)正確。", "metadata": {"timestamp": "2023-10-23 18:52:18.204795", "source": "CAP_physics-106", "explanation_source": "https://twoliunature.pixnet.net/blog/post/9069317"}, "human_evaluation": {"quality": "", "comments": ""}} |
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{"id": "CAP_physics_dev-107-45", "question": "甲、乙、丙三地位於同一條筆直的道路上,且乙地位於甲、丙之間,甲、乙二地的距離為 $S_1$,乙、丙二地的距離為 $S_2$。小明沿著道路由甲地出發經乙地到達丙地後再折返回乙地,其路線即甲→乙→丙→乙,已知此過程小明的平均速度大小為每小時 3 公里,平均速率為每小時 15 公里,則 $S_1$:$S_2$ 為下列何者?", "A": "1:1", "B": "1:5", "C": "1:4", "D": "1:2", "E": null, "F": null, "answer": "D", "explanation": "1. 本題主要是區分位移與路徑長的差異、平均速度與平均速率的差異。\n2. 乙地位於甲、丙之間,甲、乙二地的距離為S1,乙、丙二地的距離為S2\n路線 甲 → 乙 → 丙 → 乙\n位移大小:(乙-甲) S1\n路徑長:(乙-甲)+(丙-乙)+(丙-乙) S1 + S2 + S2 = S1 + 2S2\n3. 設時間為t, 平均速度大小 = 3\nS1 / t = 3 -> S1 = 3t\n平均速率 = 15\n(S1 + 2S2) / t = 15 -> S1 + 2S2= 15t\nS1 = 3t 代入 3t + 2S2= 15t -> 2S2 = 12t -> S2 = 6t\n4. S1:S2 = 3t:6t = 1:2,故(D)正確。", "metadata": {"timestamp": "2023-10-23 18:52:18.204747", "source": "CAP_physics-107", "explanation_source": "https://twoliunature.pixnet.net/blog/post/9069110-107%E5%9C%8B%E4%B8%AD%E6%9C%83%E8%80%83%E8%87%AA%E7%84%B6%E7%A7%91%E7%AC%AC45%E9%A1%8C"}, "human_evaluation": {"quality": "", "comments": ""}} |
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