求20道初中物理力学的题(难一些最好),题目好的话有追加

2025-01-23 17:39:29
推荐回答(1个)
回答1:

希望帮得上忙
用同种金属制成质量相等的实心金属球和金属盒各一个,把球放在封闭的盒内,它们恰好悬浮在水中,若把球与盒用细线相连,在水中静止时仍有1/6的体积露出水面,此时绳的拉力为20N。求:(1)金属的密度;(2)空心金属盒的体积;(3)剪断绳子,静止时金属盒露出水面的体积是多少?

2.坦克越过壕沟时,有一个简便办法:坦克上备有气袋,遇到壕沟时把气袋放下去,给气袋充气后,坦克通过壕沟就像走平地一样,设坦克的质量为4*10四次方KG,履带著地面积为5m2,当坦克的前一半履带压在气袋上时,坦克对气袋的压强是多大(设坦克前后是对称的)?

3.刘家峡水电站的水库大坝高147m,当水库水位为130m时,坝底受到的水的压强是多大?

4.一辆东风载重汽车,加上所装货物共重80000N,要匀速驶上长80m,高10m的斜坡,若摩擦阻力为5000N,汽车上坡时的牵引力至少为多大?
5.思考我们在秤量物体的时候,为什么总是选择在平衡状态的时候读数?动手实验一下,当物体突然向上运动时,在运动开始阶段,读书比平衡状态时是大还是小?

6.某同学想:地球转动速度很大,如北京地面的转动速度大约是360m/s,当人跳起来落回原地面时,地面会转过很大一段距离,自己就不会落回原地,所以要不停的跳跃就能免费周游世界。
请你考虑一下,这位同学的想法正确吗?请发表你的见解并说明理由。

7.请你想象一下,如果没有了惯性,我们世界会发生改变吗?根据这一假想,举出两个你想象的场景。答案均在下面!

1.解:(1)把球放在封闭的盒内,它们恰好悬浮在水中,则ρ水V盒=2(m盒) ①
若把球与盒用细线相连,在水中漂浮,则5/6ρ水V盒+ρ水V球=2(m盒)②
①②联立为方程组 得出ρ球=m/V球=3g/cm^3
(2)由于金属球在绳子作用下静止可知三力平衡,则20N+ρ水V球=m盒=ρ球V球,可 求出V球=1cm^3
由ρ水V盒=2m =2ρ球V球 求出V盒=6cm^3
(3)剪断绳子,有ρ水(V盒-V露)=m可求出V露=3cm^3

2.解: F=(1/2)G=(1/2)mg=1/2×4×10000Kg×9.8N/Kg=196000N
P=F/S=196000N/[1/2×5m2]=78400Pa

解析:坦克的前一半履带压在气袋上时,这是干扰条件,此时作用在气袋的压力是重力的一半,受力面积也是总面积的一样,所以压强不变.

3.解:坝底压强 P=ρgh =1.0×103`×9.8×130=1274000pa
(提示:水深是130m,与水库大坝高度无关。)

4.解:(1)假如没有摩擦力,则根据F牵*L=G得F牵*80m=80000N*10m解得:F牵=10000N ,而本题摩擦力为5000N,所以实际所需的力为10000N+5000N=15000N
(2)若按高中知识来求,则先求出"下滑力"为:80000N*(10m/80m)=10000N ,"摩擦力"加上"下滑力(即沿斜面向下的重力分量)"就是所求牵引力的最小值:
10000N+5000N=15000N

问题5.涉及到超重和失重的问题了,其实你在坐电梯的时候你能感觉到自己双腿承受的力量的变化。 称重的时候需要一个平衡状态,要么静止,要么匀速运动。
问题6.你不能这样环游世界的原因是你的惯性,你起跳的时候实际上在水平方向上你具有地球一样的速度,所以你会落到起跳点上,在车上也是一样的,一切源于惯性~
问题7.没有惯性,也许你能那样周游世界,也许你会祥光一样! 这是一个伪问题,质量是物质的基本属性必然存在的,而惯性是一切有质量物体的根本属性,没有什么能挑掉,包括光~
8、磅秤上有一个重1500N的木箱,小明站在地上,想用如图29(甲)所示的滑轮组把这个木箱提升到楼上,可是他竭尽全力也没有提起,此时磅秤的示数为40kg。于是他改变滑轮组的绕绳方法如图29(乙)所示,再去提这个木箱。当木箱匀速上升时,小明对地板的压力为100N,不计轴摩擦和绳重,取g=10N/kg。求小明的体重和提升木箱时滑轮组的机械效率。

这是图:http://blog.photo.sina.com.cn/showpic.html#url=http://static13.photo.sina.com.cn/orignal/54449990t6ed296481c0c
答案:甲图,磅秤的示数为40kg,意味着“木箱和动滑轮”受到的向上的力为1100N,因为是两股绳子,假设小明施加的拉力为F,因为“他竭尽全力也没有提起”(站在地面上的人拉绳子最多能施加的拉力大小等于人的重力),所以小明的重力大小为F.由此[(1500+G)-400]/2=F

图乙是三股绳子。小明对地板的压力是100N,那么小明对绳子施加的作用力是(F-450),由此:(1500+G)/3=F-450

接两个式子得到:G(动滑轮)=300N,F=700N(也就是人的重力)

此时的机械效率=1500/[3*(700-100)]=83.33%

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