百师联盟2022届高三第二次联考试卷和答案，知嘛答案网已经编辑汇总了百师联盟2022届高三第二次联考试卷和答案的各科答案和试卷，更多试卷答案请关注知嘛答案。
第二节参考范文Th[文]e girl's mother told Kelly that they were moving to another city. After that they could no longer run thefairy garden. She had reached an agreement[章] with the community[来] that planned to turn it into a park for local resi-dent[自]s to relax in, But her daughter did not want to leave the garden, and she had cried several times. She waseven more unwilling[知] to leave her dear fairy. Her mother wanted Kelly to meet her daughter personall[嘛]y. but shewasnt sure how to make her daughter happy.The[答] girI's mother preferred[案] Kelly to act as a fairy to say goodbye to her daughter.[网] Kelly accepted thegirls mother's request. Then they agreed on a time and a place to meet. On the day of the meeting, Kellyappe[文]ared in front of the girl dressed like a fairy. The little girl was surprised[章] and delighted[来] to see her. theyjoine[自]d hands and started dancing. She chatted excitedly[知] with Kelly for a long time, Kelly gave her many giftsthat[嘛] she liked. The girl made Kelly promise that Kelly would visit her often.
6.【答案】D解析[答]木块随圆盘一起转动[案],静摩擦力提供向心[网]力,而所需要的向心[文]力大小由物体的质量[章]、半径和角速度决定[来]。当圆盘转速增大时[自],提供的静摩擦力随[知]之而增大,当需要的[嘛]向心力大于最大静摩[答]擦力取物体开始滑动[案]。因此是否滑动与质[网]量无关本题的关键是[文]正确分析木块的受力[章],明确木块做圆周运[来]动摩擦力提供向心力[自],把握住临界条静摩[知]擦力达到最大,由牛[嘛]顿第二定律分析解答[答]解答】木块随圆盘一[案]起转动,静率擦力提[网]供向心力,由牛顿第[文]二定律得:木块所受[章]的静摩擦力f=mu[来]r,a和b的质量分[自]别是2m和m,而a[知]与转轴00的距离为[嘛]L,b与转轴的距离[答]为2L,所以开始时[案]a和b受到的摩擦力[网]是相等的;b受到的[文]静摩擦力先达到最大[章],故A错误在b的摩[来]擦力没有大前,静摩[自]擦力提供向牛顿第二[知]定律得:木块所受的[嘛]静摩擦力f=mT,[答]a和b的质量分别是[案]2m和a与转轴OO[网]的距离为L,b与转[文]轴的距离为2L,所[章]以开始时a和b受到[来]的摩擦力是相等的;[自]当b受到的静摩擦力[知]达到最大后,b受到[嘛]的摩擦力与绳子的拉[答]力的和提供向即:k[案]mg+F=m22L[网],而a的受力:f-[文]F=2mu,联立得[章]f=4m2L-km[来]g,可知二者受到的[自]摩擦力不一定相等,[知]故B错误C当b刚要[嘛]滑动时,有kmg+[答]kmg=2mo2L[案]+mu2L,解得:[网]故C错误时,此时b[文]所受摩擦力已达最受[章]摩擦力的大小为f”[来]=±m2L-kmg[自]=4m:32.L-[知]kmg故D正确故选[嘛]D
