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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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3 years ago
Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha
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Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

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Answer:

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i show the thing. i hope this helps

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A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si
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Answer:

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Explanation:

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