各位大神,帮忙看一下这道初中物理题,在线等

初中
2024-11-28 23:00:46
推荐回答(3个)
回答1:

1)500N
首先F1*S=W
所以 F1*3*0.2*t=W
从图中带数(2,900)
F1=750N
分析受力
杠杆A端受力为
G求+F1
杠杆B端受力为
800-300*1=500N
所以根据杠杆定理
(G求+750)*OA=500*OB
解得G求=500N

2)
杠杆A端受力为
G求+F2=500+F2
杠杆B端受力为
800-200*1=600N
所以根据杠杆定理
(500+F2)*OA=600*OB
F2=1000N
v1:v2=2/3
v2=0.3
P=F2*S2/t2
=1000*3*0.3
=900
3)
F1拉力时 设1秒钟 有效功为
W有1=G1*0.2
W1=F1*S=3*750*0.2=450
n1=G1*0.2/450=G1/2250
F2拉力时,设1s有效功为
W有2=2750*0.3=825
W2=F2*3*0.3=1000*3*0.3=900
n2=825/900=11/14
所以n1:n2=G1/2250*14/11=7G1/12375
(PS:按照道理第三问求的的应该是这个样子,但是不是说忽略所有绳子的摩擦什么的,那么机械效率是不意味着是一样了?这个你需要自己再考虑一下)

回答2:

我只是路过而已 虽然不懂楼主在说什么 但好像很厉害的样子

百 练成仙
度 过天劫
知 己知彼
道 行高深
是 神是鬼
垃 个碑墓
圾 地转世

回答3:

楼主你等一下,我正在解

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