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Sergio039 [100]
3 years ago
13

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

the wing is 199 m/s. The density of the air is 1.29 kg/m3. What is the lifting force on a wing of area 27.0 m2?
Physics
1 answer:
grin007 [14]3 years ago
6 0
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

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A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have
Doss [256]

Answer:

2000\; {\rm cm^{3}}.

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let m(\text{ball}) denote the mass of this ball. Let m(\text{water}) denote the mass of water that this ball has displaced.

Let g denote the gravitational field strength. The weight of this ball would be m(\text{ball}) \, g. Likewise, the weight of water displaced would be m(\text{water})\, g.

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

\text{buoyancy} \ge m(\text{ball})\, g.

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

\text{buoyancy} = m(\text{water}) \, g.

Therefore:

m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g.

m(\text{water}) \ge m(\text{ball}).

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let \rho(\text{water}) denote the density of water. The volume of water that this ball should displace would be:

\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \end{aligned}.

Given that m(\text{ball}) = 2000\; {\rm g} while \rho = 1.00\; {\rm g\cdot cm^{-3}}:

\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}.

In other words, for this ball to stay afloat, at least 2000\; {\rm cm^{3}} of the volume of this ball should be under water. Therefore, the volume of this ball should be at least 2000\; {\rm cm^{3}}\!.

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