Skip to content

Categories:

範例:牛頓第二定律與空氣阻力[101學測]

由離地相同高度處,於同一瞬間,使甲球與乙球自靜止狀態開始落下,兩球在抵達地面前,除重力外,只受到來自空氣阻力F的作用,此阻力與球的下墜速度v成正比,即F=-kv(k>0),且兩球的比例常數k完全相同,圖所示為兩球的速度-時間關係圖。(1)若甲球與乙球的質量分別為m1與m2,則下列敘述何者正確?(A) m1=m2,且兩球同時抵達地面 (B) m2>m1,且乙球先抵達地面 (C) m2<m1,且乙球先抵達地面 (D) m2<m1,且兩球同時抵達地面 (E) m2>m1,且甲球先抵達地面

(2)若已知甲球質量為0.2公斤,落下過程中重力加速度恆為10 公尺/秒2,則比例常數k值約為多少公斤/秒?(A)0.1 (B)0.2 (C)4 (D)10 (E)40

相關文章:

Posted in 歷屆學測(推甄)考題, 歷屆學測試題, 牛頓運動定律.

Tagged with , , , .


0 Responses

Stay in touch with the conversation, subscribe to the RSS feed for comments on this post.

You must be logged in to post a comment.



33 visitors online now
9 guests, 9 bots, 15 members
All time: 479 at 08-18-2020 08:10 pm UTC
Max visitors today: 80 at 03:20 am UTC
This month: 222 at 07-01-2024 11:54 pm UTC
This year: 222 at 07-01-2024 11:54 pm UTC