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liubo4ka [24]
3 years ago
7

How is buoyancy used in the real world?​

Physics
2 answers:
JulsSmile [24]3 years ago
7 0

Answer:

to mack boats float

Explanation:

without it the boats will sink to the bottom

seraphim [82]3 years ago
4 0

-- Buoyancy is used to bring crude oil from Saudi Arabi to refineries in the US. The oil is loaded into tanks that are part of supertanker ships. Buoyancy is created by their ship-shape, so they float on water, and they can be dragged across the ocean on top of the water.

-- Buoyancy is used for a cheap thrill on the weekend.  People drive out in the country and find a place where they can buy a ride under a hot-air balloon.  The balloon is filled with hot air from a propane burner under its opening, and then it rises up off the ground because of its buoyancy in cool air.

-- Buoyancy is used cleverly by factories, to get rid of their gaseous and particulate wastes.  The wastes are heated, and then blown into tall "smokestacks" connected to the factory.  Then they rise because of the buoyancy created by hot stuff in cool air.  When they leave the top of the smokestack, they keep rising for a while.  Then, weather systems blow them away from the factory, over into other people's neighborhoods, where they finally sink to the ground in places where nobody knows where they came from.

-- Buoyancy is used in teapots, coffee makers, and lava lamps.  A heat source is placed under the bottom of the container. Heat is conducted through the bottom of the container, and it heats the fluid that's in contact with the inside of the container on the bottom. The hot fluid rises to the top of the container, because of the buoyancy created by hot stuff surrounded by cool stuff.  At the same time, cool stuff flows in to take its place at the bottom.  

In pots and pans and tanks where buoyancy is used this way, the motion of the fluids up and down and around is called "convection".

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Newton is denoted by N.

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If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
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Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

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V2 = ?

Formula

                          P1V1 = P2V2

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The greatest obstacle to developing solar energy is _____. lack of sunlight at high latitudes lack of sunlight at night lack of
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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j
AlekseyPX

1) Fundamental units of \Psi are [\frac{mol}{m\cdot s^2}]

2) Fundamental units of \Phi are [\frac{mol}{m^3}]

Explanation:

The equation for the variable \rho is

\rho =\frac{2\gamma \Phi+\Psi}{rg}

where we have:

\rho measured in [\frac{mol}{ft^3}]

\gamma measured in [\frac{J}{kg}]

r measured in [in]

g measured in [\frac{m}{s^2}]

We can re-write the equation as

\rho rg = 2\gamma \Phi + \Psi

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, \Psi must have the same units of \rho r g. So,

[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of \Psi are

[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]

2)

The term 2\gamma \Phi must have the same units of \Psi in order to be added to it. Therefore,

[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]

We also know that the units of \gamma are [\frac{J}{kg}], therefore

[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]

And so, the fundamental units of \Phi are

[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]

However, the Joules can be written as

[J]=[kg][\frac{m^2}{s^2}]

Therefore

[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]

#LearnwithBrainly

7 0
3 years ago
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