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Masteriza [31]
3 years ago
11

A tank has a gate that automatically opens if the water levelhis high enough. The gate has a squarecross section of side1m and c

an rotate about an axis going through the point O. Calculate the minimumwater level that would trigger the opening of the floodgate.

Physics
1 answer:
umka2103 [35]3 years ago
3 0

Answer:

The gate will open if the height of water is equal to or more than 0.337m.

Explanation:

From the diagram attached, (as seen from the reference question found on google)

The forces are given as

Force on OA

F_1=P A_1\\F_1=\rho g \bar{h} A_{OA}

Here

  • ρ  is the density of water.
  • g is the gravitational acceleration constant
  • \bar{h} is the equivalent height given as

         \bar{h}=h+\frac{0.6}{2}\\\bar{h}=h+0.3

  • A_{OA} is the area of the OA part of the door which is calculated as follows:

       A_{OA}=L\times W\\A_{OA}=1\times 0.6\\A_{OA}=0.6 m^2

The  Force is given as

F_1=0.6\rho g[h+0.3]

Force on OB

F_2=P A_2\\F_2=\rho g \bar{h} A_{OB}

Here

  • ρ  is the density of water.
  • g is the gravitational acceleration constant
  • \bar{h} is the equivalent height given as

         \bar{h}=h+0.6+\frac{0.4}{2}\\\bar{h}=h+0.8

  • A_{OA} is the area of the OB part of the door which is calculated as follows:

       A_{OB}=L\times W\\A_{OB}=1\times 0.4\\A_{OB}=0.4 m^2

The  Force is given as

F_2=0.4\rho g[h+0.8]

Now the moment arms are given as

\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}

\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}

Taking moment about the point O as zero

F_1(h+0.6-\bar{y}_a)=F_2(\bar{y}_b-h+0.6)\\F_1(h+0.6-h-0.3-\frac{0.03}{h+0.3})=F_2(h+0.8+\frac{0.0133}{h+0.8}-h-0.6)\\F_1(0.3-\frac{0.03}{h+0.3})=F_2(0.2+\frac{0.0133}{h+0.8})\\0.6\rho g[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m

So the gate will open if the height of water is equal to or more than 0.337m.

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Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

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Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

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Explanation:

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a)

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Xc= \frac{1}{2\pi fC}

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z=\sqrt{ R^2 + (Xl - Xc)^2}

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