physics - fluid
2009-09-15 4:49 am
An open tank holds water 1.25m deep. If a small hole of cross-section area 3cm^2 is made at the bottom of the tank , calculate the mass of the water per second initially flowing out of the hole.(g=10m/s^2, density of water=1000kg/m^3)
回答 (2)
2009-09-15 4:17 pm
✔ 最佳答案
By Bernoulli's Principle,P0 + pgh + 1/2 pu2 = P0 + pgh0 + 1/2 pv2
pgh = 1/2 pv2
v = sqrt(2gh)
v = sqrt[2(10)(1.25)] = 5 ms-1
Therefore, the water flows out with the velocity of 5 ms-1
With the cross-section area 3 cm2 = 3 X 10-4 m2
Volume of water flowing out per second
= (5)(3 X 10-4)
= 15 X 10-4 m3
So, the mass of water flowing out per second
= 15 X 10-4 X 1000
= 1.5 kg
參考: Physics king
2009-09-19 4:59 am
By Bernoulli's Principle,
P0 + pgh + 1/2 pu2 = P0 + pgh0 + 1/2 pv2
pgh = 1/2 pv2
v = sqrt(2gh)
v = sqrt[2(10)(1.25)] = 5 ms-1
Therefore, the water flows out with the velocity of 5 ms-1
With the cross-section area 3 cm2 = 3 X 10-4 m2
Volume of water flowing out per second
= (5)(3 X 10-4)
= 15 X 10-4 m3
So, the mass of water flowing out per second
= 15 X 10-4 X 1000
= 1.5 kg
P0 + pgh + 1/2 pu2 = P0 + pgh0 + 1/2 pv2
pgh = 1/2 pv2
v = sqrt(2gh)
v = sqrt[2(10)(1.25)] = 5 ms-1
Therefore, the water flows out with the velocity of 5 ms-1
With the cross-section area 3 cm2 = 3 X 10-4 m2
Volume of water flowing out per second
= (5)(3 X 10-4)
= 15 X 10-4 m3
So, the mass of water flowing out per second
= 15 X 10-4 X 1000
= 1.5 kg
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