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Doss [256]
3 years ago
14

A container is filled to a depth of 22.0 cm with water. On top of the water floats a 28.0-cm-thick layer of oil with specific gr

avity 0.500. What is the absolute pressure at the bottom of the container? 1 Pa
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:104.853 kPa

Explanation:

Given

Depth of water h_2=22 cm

on the top of water a layer of h_1=28 cm oil layer is present

specific gravity of oil \rho _o=0.5\rho _w

Assuming Atmospheric Pressure to be 101.325 KPa

We know Pressure due to Pressure Difference is given by

\Delta P=\rho \times g\times h

\Delta P=\rho _0\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\rho _w\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\times 1000 \times 9.8\times 0.28+1000\times 9.8\times 0.22

\Delta P=1.372\times 10^3+2.156\times 10^3

\Delta P=3.528 kPa

Absolute\ Pressure=Atmospheric\ Pressure+\Delta P=101.325+3.528

Absolute\ Pressure =101.325+3.528=104.853 kPa

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Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate.

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4 years ago
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The free body diagram represents Silly Sally hanging from a trapeze bar. Sally weighs 660 Newtons. What is the force in each of
Y_Kistochka [10]
A) 330 N

Explanation: Her weight must be evenly distributed between the chains (assuming they are at an even level), so you divide 660 by 2 and get 330
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4 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
3 years ago
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Svetlanka [38]

Answer:

An electrical current

Explanation:

An electrical current

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