3个回答
展开全部
(1)设拉力为F,滑轮G1,中午G2,则F = (G1+G2)/2 = (300 + 1200)/2N = 750N
(2)设S = 4米,η = G2S/(G1+G2)S = G2/(G1+G2)= 1200N/(1200N + 300N)= 80%
(3)工人做功 W = (G1+G2)S = (1200N + 300N) × 4m = 6000J,
所以功率P = W/t = 6000J/20s = 300w
(2)设S = 4米,η = G2S/(G1+G2)S = G2/(G1+G2)= 1200N/(1200N + 300N)= 80%
(3)工人做功 W = (G1+G2)S = (1200N + 300N) × 4m = 6000J,
所以功率P = W/t = 6000J/20s = 300w
更多追问追答
追问
为什么G1+G2除以二呢?
不是三段绳子吗?
为什么G1+G2除以二呢?
不是三段绳子吗?
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询