一道有关工作效率的数学问题,好困惑,不知道怎么求。
TapAcanfillawatertankin25minutes,andTapBcanfillthesametankin20minutes.(a)Ifthevolumeo...
Tap A can fill a water tank in 25 minutes,and Tap B can fill the same tank in 20 minutes.
(a)If the volume of the tank is 5000cm^3,find the rate of flow of water(in cm^3/min)for
(i)Tap A,and
(ii)Tap B,
(b)Both taps(Tap A and Tap B )are used to fill up the tank at the same time.
(i)Calculate the total volume of water in the water tank after 10 minutes.
(ii)Will the water overflow if both Tap A and Tap B are turn on for 10 minutes. 展开
(a)If the volume of the tank is 5000cm^3,find the rate of flow of water(in cm^3/min)for
(i)Tap A,and
(ii)Tap B,
(b)Both taps(Tap A and Tap B )are used to fill up the tank at the same time.
(i)Calculate the total volume of water in the water tank after 10 minutes.
(ii)Will the water overflow if both Tap A and Tap B are turn on for 10 minutes. 展开
1个回答
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【题目】
单开龙头A用25分钟可以灌满水池,而单开龙头B灌满同样的水池要用20分钟。
(a)若水池容积为5000立方厘米,求出
(i)龙头A的流速,以及
(ii)龙头B的流速;
(b)若同时打开龙头A和龙头B,
(i)计算10分钟后水池中水的总体积;
(ii)10分钟后水会溢出吗?
【解】
(a)
(i)龙头A水流速度=5000/25=200(cm^3/min)
(ii)龙头B水流速度=5000/20=250(cm^3/min)
(b)
(i)10分钟后池中水体积=(200+250)×10=4500(cm^3)
(ii)∵4500<5000
∴水不会溢出
单开龙头A用25分钟可以灌满水池,而单开龙头B灌满同样的水池要用20分钟。
(a)若水池容积为5000立方厘米,求出
(i)龙头A的流速,以及
(ii)龙头B的流速;
(b)若同时打开龙头A和龙头B,
(i)计算10分钟后水池中水的总体积;
(ii)10分钟后水会溢出吗?
【解】
(a)
(i)龙头A水流速度=5000/25=200(cm^3/min)
(ii)龙头B水流速度=5000/20=250(cm^3/min)
(b)
(i)10分钟后池中水体积=(200+250)×10=4500(cm^3)
(ii)∵4500<5000
∴水不会溢出
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