【物理】problem on fluids


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A cylindrical container with a cross-sectional area of 57.3 cm2 holds a fluid of density 840 kg/m3. At the bottom of the container the pressure is 110 kPa.

(a) What is the depth of the fluid?

wrong check mark

Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m

(b) Find the pressure at the bottom of the container after an additional 1.50 10-3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.

kPa

For (a) can someone calculate again for me because I kept getting the same answer. I can't find the error in my calculation. If you have time give me the answer for (b) too, lol. I'm lazy.....

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為了方便,單位用MKS制,計算省略單位。

1.840*h*9.81=110*1000

h=13.3(m)

2.1.50*10^-3/(57.3*10^-4)=0.262(m)

13.3+0.262=13.6(m)

840*13.6*9.81=1.12*10^5(Pa)

也就是112kPa

不知和樓主算的有無不同?

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那為什麼算出 0.262 之後

又要拿水下去算

在加那個高度?

2.1.50*10^-3/(57.3*10^-4)=0.262(m)

13.3+0.262=13.6(m)

840*13.6*9.81=1.12*10^5(Pa)

第一題中算出與該壓力等量的液體高度(類似mmHg、cmHg之類)

接著他把新增加的高度算出來(第一行)

然後加上原有的高度(第二行)

最後換算成壓力(第三行)

應該是這樣子

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