一道物理题 速度在线等

2024-12-24 18:26:54
推荐回答(5个)
回答1:

v₁=20
v₂=10 a₂= 0.5m/s^2
S1=100m第一步,我们先算出货车从减速到速度为零是需要的时间,不难算出时间t=20s第二步,我们再算出货车从减速直到速度为零这段时间又走了多少路程:S2=v₁t-1/2 a₂ t2 ,代入相关数值可解得S2=100m,注意此时货车已经速度为零第三步,我们来看小汽车,小汽车与货车是同时减速的,当货车速度为零时小汽车速度要保证也是零才可以,这样它们才不会撞上,而且小汽车此时刚好追上货车,这时候求出来的加速度是最小的,即是题目要求的,这是一个临界点,要抓住,接下来就好办了。小汽车走的路程S= S1+s2=200m,此时速度已经为零,所以v₁^2=2as,解得a=1m/s^2,所以加速度至少为1m/s^2。

回答2:


w=Pt=60×0.5=30KW
p=F×v F=P/v=60000÷10=6000N
因为车是水平路面上匀速行驶 所以F阻=F牵=6000N

回答3:

第一题,小球本来就是向前运动的,由于惯性,所以会继续向前,张力会增大,小球此时相当于位于圆周运动的最低点,圆周运动向心力由拉力提供F-mg=mV2/r,即F=mg+mV2/r,原来F=mg,所以F增大了,选AC。第二题,(1)最小速度对应的是在最高点绳子拉力为零,重力提供向心力mg=mV2/L,所以V=2.5m/s。(2)F+mg=mV2/L,即可求出F=9.76N,绳子断了,物体具有水平方向的速度和受到向下的重力,所以做平抛运动。

回答4:

答案:(1)Tap=10√3+50牛顿;Tbp=50√3-10牛顿,方向斜右上方,与坚直夹角30度。
(2)Tbp=100-50√3牛顿
Tap=100√3+50牛顿
(由于百度不支持上下标以及数学模板的运算,所以只能直接给出答案,非常抱欠了。)
建议百度能在数学支持这一方面做一些改进。这应该不会很难吧!

回答5:

1由图可知温度从10℃升至30℃时发热体电阻减小,温度从10℃升至100℃时发热体电阻增大,由P=U²/R可知电压一定,从10℃升至30℃时电功率增大,温度从10℃升至100℃时电功率减小。2.发热体的温度为30℃时R=2000欧
P=U²/R=(220v)²/2000欧=24.2w
3.R=U²/P=(220v)²/4,84W=10000欧=10千欧
由图可知温度60℃

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