第3 條:
A box moves at a uniform velocity of 2ms-1 on a frictionless horizontal surface. sand falls into the box with negligible speed at a rate of 90kg per minute. what horizontal force is required to keep the box moving uniformly at 2ms-1??
A. 0N
B. 3N
C. 6N
D. 90N
E. 180N
Which is correct ?? WHY?
急....93年 phy al mc 問題
2009-11-15 8:18 pm
回答 (2)
2009-11-15 8:31 pm
✔ 最佳答案
Rate of increase of mass of box = 90/60 kg/shence, rate of increase of momentum of box = (90/60) x 2 kg-m/s2
Since force = rate of change of momentum
i.e. force required = (90/60) x 2 N = 3 N
2009-11-15 11:49 pm
Rate of falling of sand
= 90 kg/min
= 1.5 kg /s
Initial Momentum of the box
= mv
= 2m
After one second, Momentum
= 2(m+1.5)
= 2m +3
Hence, increase in Mometum = 3(kgv/s)
Force F = change in Mometum / t
Since t = 1(s),
F
= change in Momentum
= 3(N)
Therefore, answer is B
= 90 kg/min
= 1.5 kg /s
Initial Momentum of the box
= mv
= 2m
After one second, Momentum
= 2(m+1.5)
= 2m +3
Hence, increase in Mometum = 3(kgv/s)
Force F = change in Mometum / t
Since t = 1(s),
F
= change in Momentum
= 3(N)
Therefore, answer is B
參考: /
收錄日期: 2021-04-29 17:31:16
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