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3 years ago

What happens to the density of gas when it compressed

Physics

1 answer:

Stella [2.4K]3 years ago

3 0

When a gas is compressed, its particles are pushed closer together so it is more dense.

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got 2 kettle .One got hot water only and the other one got hot water+tea .which kettle got higher boiling points?

FromTheMoon [43]

Answer:

The kettle with hot water only.

Explanation:

Adding tea to hot water reduces the attractive forces between water molecules therefore the water molecules in hot water plus tea will move away from each other faster hence hastening the boiling point.

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4 years ago

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An object in a(n) __________ orbit in the Solar System will remain in its orbit forever. An object in a(n) __________ orbit will

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bound; unbound

Explanation:

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3 years ago

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

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