A.質(zhì)點(diǎn)此時(shí)的振動(dòng)方向沿軸負(fù)方向 查看更多

 

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沿x軸正方向傳播的一列簡(jiǎn)諧橫波,波第一次傳播到x=10m處時(shí)形成如圖所示的波形,則(   )

A.此時(shí)刻x=3m處質(zhì)點(diǎn)正沿y軸正方向運(yùn)動(dòng)

B.此時(shí)刻x=6m處質(zhì)點(diǎn)的速度為零

C.此時(shí)刻x=8m處質(zhì)點(diǎn)的已經(jīng)運(yùn)動(dòng)了1/4周期

D.x=8m處質(zhì)點(diǎn)的開(kāi)始振動(dòng)的方向沿y軸負(fù)方向

 

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沿x軸正方向傳播的一列簡(jiǎn)諧橫波,波第一次傳播到x=10m處時(shí)形成如圖所示的波形,則(  )
A.此時(shí)刻x=3m處質(zhì)點(diǎn)正沿y軸正方向運(yùn)動(dòng)
B.此時(shí)刻x=6m處質(zhì)點(diǎn)的速度為零
C.此時(shí)刻x=8m處質(zhì)點(diǎn)的已經(jīng)運(yùn)動(dòng)了1/4周期
D.x=8m處質(zhì)點(diǎn)的開(kāi)始振動(dòng)的方向沿y軸負(fù)方向

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沿x軸正方向傳播的一列簡(jiǎn)諧橫波,波第一次傳播到x=10m處時(shí)形成如圖所示的波形,則(   )

       A.此時(shí)刻x=3m處質(zhì)點(diǎn)正沿y軸正方向運(yùn)動(dòng)

       B.此時(shí)刻x=6m處質(zhì)點(diǎn)的速度為零

       C.此時(shí)刻x=8m處質(zhì)點(diǎn)的已經(jīng)運(yùn)動(dòng)了1/4周期

       D.x=8m處質(zhì)點(diǎn)的開(kāi)始振動(dòng)的方向沿y軸負(fù)方向

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沿x軸正方向傳播的一列簡(jiǎn)諧橫波,波第一次傳播到x=10m處時(shí)形成如圖所示的波形,則(   )

A.此時(shí)刻x=3m處質(zhì)點(diǎn)正沿y軸正方向運(yùn)動(dòng)

B.此時(shí)刻x=6m處質(zhì)點(diǎn)的速度為零

C.此時(shí)刻x=8m處質(zhì)點(diǎn)的已經(jīng)運(yùn)動(dòng)了1/4周期

D.x=8m處質(zhì)點(diǎn)的開(kāi)始振動(dòng)的方向沿y軸負(fù)方向

 

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沿x軸正方向傳播的一列簡(jiǎn)諧橫波,波第一次傳播到x=10m處時(shí)形成如圖所示的波形,則(  )

A.此時(shí)刻x=3m處質(zhì)點(diǎn)正沿y軸正方向運(yùn)動(dòng)
B.此時(shí)刻x=6m處質(zhì)點(diǎn)的速度為零
C.此時(shí)刻x=8m處質(zhì)點(diǎn)的已經(jīng)運(yùn)動(dòng)了1/4周期
D.x=8m處質(zhì)點(diǎn)的開(kāi)始振動(dòng)的方向沿y軸負(fù)方向

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一、本題共14小題.每小題3分。共42分.全部選對(duì)的得3分,選對(duì)但不全的得2分.有選錯(cuò)或不答的得0分.

1.C    2.AD    3.BD    4.B    5.A B    6.AC    7.D   8.BD    9.C    l0.B    11.D    12.BC    13.B    14.D

二、實(shí)驗(yàn)題:本題共2小題,共14分.

15.(6分)

(1)BD(4分)

(2)左(2分)

16.(8分)

(1)D,F(xiàn)(每空1分)

(2)(3分)

    (3)(3分)

三、本題共4小題,共44分。

17.(10分)

(1)設(shè)排球第一次落至地面經(jīng)歷的時(shí)間為,第一次離開(kāi)地面至反彈到最高點(diǎn)經(jīng)歷    的時(shí)間為,則有

             

                                                                                                            (2分)

解得                                                                            (2分)

所以,排球與地面的作用時(shí)間為                               (1分)

(2)設(shè)地面對(duì)排球的平均作用力為,選排球?yàn)檠芯繉?duì)象,以向上的方向?yàn)檎较,則在整個(gè)過(guò)程中,由動(dòng)量定理有                                                       (3分)

代入數(shù)據(jù)解得:                                                                        (1分)

根據(jù)牛頓第三定律,排球?qū)Φ孛娴钠骄饔昧?sub>                          (1分)

18.(11分)

設(shè)離子的質(zhì)量為、電荷量為,磁場(chǎng)的磁感應(yīng)強(qiáng)度為、所在區(qū)域的半徑為,離子加速后獲得的速度為.當(dāng)電壓為時(shí),由動(dòng)能定理有

                                           ①                                                   (2分)

在磁場(chǎng)中,離子做勻速圓周運(yùn)動(dòng)(見(jiàn)右圖),

由牛頓定律可知

                                                  ②   (2分)

由①②式得                           ③           (1分)

其中,                    ④                                                     (1分)

當(dāng)電壓為時(shí),離子打在

同理有                                ⑤                                                     (2分)

其中,                                          ⑥                                                     (1分)

由③④⑤⑥可解得                                                                              (2分)

19.(11分)

(1)小球沿方向作豎直上運(yùn)動(dòng),沿方向作初速為零的勻加速運(yùn)動(dòng),設(shè)點(diǎn)坐標(biāo)    為(),勻強(qiáng)電場(chǎng)的場(chǎng)強(qiáng)為,小球沿方向的加速度為,由牛頓第二定律得:

                                         ①                                                 (1分)

由運(yùn)動(dòng)規(guī)律,得:

                                        ②                                              (1分)

由動(dòng)量定義,得:

                                                              ③                                                     (1分)

             設(shè)經(jīng)時(shí)間,小球從原點(diǎn)運(yùn)動(dòng)到Q點(diǎn),則

                                                    ④                                                  (1分)

                                       ⑤                                                   (1分)

由①②③④⑤得

                                                                           (1分)

(2)小球從拋出到落回軸,經(jīng)時(shí)間,在方向位移

                                                                         (2分)

電場(chǎng)力對(duì)小球作功

                                                            (2分)

    電場(chǎng)力做正功,電勢(shì)能減少。                                                    (1分)

20.(12分)

(1)設(shè)物塊與板間的動(dòng)摩擦因數(shù)為,物塊在板上滑行的時(shí)問(wèn)為,對(duì)板應(yīng)用動(dòng)量定

  理得

                                   ①                                                     (1分)

設(shè)在此過(guò)程中物塊前進(jìn)位移為,板前進(jìn)位移為,

                                        ②                                                     (1分)

                                          ③                                                     (1分)

                                      ④                                                     (1分)

              由①②③④得:                                                                       (2分)

      故物塊與板間的摩擦因數(shù)為

(2)設(shè)板與桌面間的動(dòng)摩擦因數(shù)為,物塊在板上滑行的時(shí)間為,則對(duì)板應(yīng)用動(dòng)量

  定理得

                                                                       (2分)

又設(shè)物塊從板的左端運(yùn)動(dòng)到右端的時(shí)間為,則

                                                                                                    (2分)

為了使物塊能到達(dá)板的右端,必須滿足:,

                                                                                                 (1分)

所以為了使物塊能到達(dá)板的右端,板與桌面間的摩擦因數(shù)

 

 


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