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Fittoniya [83]
3 years ago
5

What is the relationship between the acceleration of an object change in relation to its mass

Physics
1 answer:
yarga [219]3 years ago
3 0

The relationship between mass and acceleration is an inverse proportionality

Explanation:

The relationship between the acceleration of an object and its mass is given by Newton's second law, which states that:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

From the equation, we notice that if the force on the object is kept constant, then the mass and the acceleration are inversely proportional to each other. This means that:

  • If the mass of the object is increased, its acceleration will decrease
  • If the mass of the object is decreased, its acceleration will increase

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t
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Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

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We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

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From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

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Answer:

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