A ________ is a collection of related data, equivalent to an electronic file cabinet.

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

umka2103 [35]

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.

3 0

3 years ago

The object distance for a concave lens is 8.0 cm, and the image distance is 12.0 cm. The height of the object is 4.0 cm. What is

swat32

The answer for this question, If I am correct, should be answer "D".

3 0

3 years ago

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What force must be applied to move a 251 kg rock on a pavement like surface

Alchen [17]

If the rock is just sitting there and you want to SLIDE it, then you have to push it with a force of at least

(251 kg) x (9.8 m/s²) x (μ) =

(2,459 Newtons) x (the coefficient of static friction on that surface)

4 0

3 years ago

A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming

worty [1.4K]

d. 49.0 m/s

this is your answer. ....OK. ..

please mark me as brainliest. .....

follow me.....my friend. .....

3 0

3 years ago

Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters.

VashaNatasha [74]

What a delightful little problem !
Here's how I see it:

When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.

Then, when 'C' is touched to 'B', charge flows to it until the two of them are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.

To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.

The force between any two charges is:

F = (a constant) x (one charge) x (the other one) / (the distance between them)².

For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.

The original force between them was 3 = (some constant) x (1 charge) x (1 charge).

The new force between them is F = (the same constant) x (1/2) x (3/4) .

Divide the first equation by the second one, and you have a proportion:

3 / F = 1 / ( 1/2 x 3/4 )

Cross-multiply this proportion:

3 (1/2 x 3/4) = F

F = 3/2 x 3/4 = 9/8 = 1.125 newton.

That's my story, and I'm sticking to it.

5 0

3 years ago