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stellarik [79]
3 years ago
8

Determine the pressure point in a liquid container within a pool. The height of the liquid column is 9 m, The density of the liq

uid is 997 kg/m^3, and gravity is 9.8 m/s^2.
Physics
1 answer:
7nadin3 [17]3 years ago
6 0

Absolute pressure: 1.89\cdot 10^5 Pa

Explanation:

The total pressure at a certain depth of a liquid is given by

p=p_0 + \rho g h

where

p_0 = 1.01\cdot 10^5 Pa is the atmospheric pressure

\rho is the density of the liquid

g is the acceleration of gravity

h is the depth of the liquid

For the liquid in this problem, we have

\rho=997 kg/m^3 is the density

g=9.8 m/s^2

h = 9 m is the column of liquid

Substituting, we find

p=1.01\cdot 10^5 + (997)(9.8)(9)=1.89\cdot 10^5 Pa

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

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The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle
Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

         Re = ρ v D /μ

The units of each term are

       ρ = [kg / m³]

       v = [m / s]

      D = [m]

      μ = [Pa s]

The pressure

      Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

      μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

      Re = [kg / m³] [m / s] [m] / [kg / m s]

      Re = [kg / m s] / [m s / kg]

      RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity       μ = 11.2 10⁻⁶ Pa s

the density  ρ = 0.656 kg / m³

       D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

       Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

       Re = 1.19 10⁴

4 0
3 years ago
One mole of water is equivalent to 18 grams of water. A glass of water has a mass of 200 g. How many moles of water is this?
Black_prince [1.1K]
200g*1 mole/ 18g=11.1 moles There are 11.1 moles of water. 
4 0
3 years ago
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Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

f(x) = 8x^3-3x^2-5x^8

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0
4 years ago
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A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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