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Oliga [24]
3 years ago
15

A gauge is reading the pressure at the bottom of a river, at a depth of 6 m. Would the reading be greater or smaller than the re

ading at the bottom of a lake at the same depth? You must provide a clear explanation for full credit.
Physics
1 answer:
Alex787 [66]3 years ago
5 0

Answer:

The pressure at the bottom of the river is less than that at the bottom of the lake.

Explanation:

From Bernoulli's equation, the pressure difference is given by

ΔP = ρgΔh + ρ(v₂² - v₁²)/2 where ρ = density of water, g = acceleration due to gravity, Δh = depth, v₁ = velocity at top, v₂ = velocity at bottom

For the lake, v₁ = v₂, since the velocity at the top and bottom are the same. So,

ΔP₁ = ρgΔh + ρ(v₁² - v₁²)/2 = ρgΔh + 0 = ρgΔh

P₂ - P₁ = ρgΔh

P₂ = P₁ + ρgΔh

For the river, v₁ < v₂, since the velocity at the top of the river is greater than at the bottom.

So,

ΔP₂ = ρgΔh + ρ(v₂² - v₁²)/2.

Since  v₁ < v₂, ρ(v₂² - v₁²)/2 will be negative,

So,

ΔP₂ = ρgΔh - ρ(v₂² - v₁²)/2.

Since ρ(v₂² - v₁²)/2 is negative, making ΔP less than that in the lake.

So, ΔP₂ = ΔP₁ -  ρ(v₂² - v₁²)/2.

ΔP₂ = P₃ - P₁

P₃ - P₁ = P₂ - P₁ -  ρ(v₂² - v₁²)/2.

P₃ = P₂ - ρ(v₂² - v₁²)/2.

where P₃ = pressure at bottom of the river and P₂ = pressure at bottom of the lake and P₁ = atmospheric pressure at top of river and lake respectively.

Since the factor ρ(v₂² - v₁²)/2 is removed from the pressure at the bottom of the lake, the pressure at the bottom of the river is therefore less than that at the bottom of the lake.

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Answer:

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A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 ​ =50m/s hits an iron bloc
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Explanation:

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½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

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The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb  + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

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