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kramer
2 years ago
6

Plz help........................

Physics
1 answer:
hoa [83]2 years ago
7 0

Answer:

The correct wording is

  1. Pressure increases with the depth of the fluid.
  2. A plane's engines produce thrust to push the plane forward.
  3. A fluid can be a liquid or a gas.
  4. A hydraulic device uses Pascal's principle to lift or move objects.
  5. lift is the upward force exerted on objects by fluids.

Explanation:

1. As you go deeper into a fluid,<em> there is more of it on top of you; </em>therefore, the pressure excreted on you is greater.

2. A plane's engines pushes the air in opposite direction, which according to newton's third law, produces necessary force to move the plane forward.

3.  <em>A fluid has no fixed shape,</em> and it deforms under the influence of external forces applied—liquid and gases fit into this definition.

4. Pascal's principle <em>says that pressure applied on one region of the fluid must equal pressure transmitted to another region of the same fluid</em>. This principle is used in a hydraulic device to exert forces on fluids to lift objects that would otherwise be difficult to move.

5. By definition, the upward force exerted by the fluids on objects is the lift.

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A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where
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Answer:

a) 90 kJ

b) 230.26 kJ

Explanation:

The pressure at the first point  P_{1}= 10 bar —> 10 x 102 = 1020 kPa

The volume at the first point  V_{1}= 0.1 m^3  

The pressure at the second point P_{2}= 1 bar —> 1 x 102 = 102 kPa

The volume at the second point V_{2} = 1 m^3  

Process A.

constant volume V = C from point (1) to P = 10 bar.

Constant pressure P = C to the point (2).  

Process B.

The relation of the process is PV = C  

Required  

For process A & B

(a) Sketch the process on P-V coordinates

(b) Evaluate the work W in kJ.  

Assumption  

Quasi-equilibrium process

Kinetic and potential effect can be ignored.  

Solution

For process A.

V=C  

There is no change in volume then

W_{a(1)}= 0\\P=10^{2}

The work is defined by  

W_{a(2)}=\int\limits^V_V {P} \, dV

W_{A(2)} =║10^{2} V║limit 1--0.1

W_{A(2)} = 90 kJ

Process B  

PV=C  

By substituting with point (1) C = 10^2 x 1= 10^2  

The work is defined by

W_{b}=\int\limits^V_V {P} \, dV\\P=10^{2} V^{-1}\\ W_{b}=\int\limits^V_V {10^{2} V^{-1}} \, dV\\\\

W_{A(2)} = ║10^{2} ln(V)║limit 1--0.1

         =230.26 kJ

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