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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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A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
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3 years ago
You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
Neporo4naja [7]

Answer:

The distance is r_2  =  0.24 \  m

Explanation:

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       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

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      d =  \frac{P}{A}

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    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

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Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

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        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

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7 0
3 years ago
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