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77julia77 [94]
3 years ago
13

How much force is generated when a 200 kg block is accelerated at a rate of 25 m/s?

Physics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

F= 5000 N

Explanation:

This problem is related to force on a body, and to tackle it we need to apply newtons first law of motion which states that "a body will continue to be at rest or uniform motion except acted upon by an external force greater than the force keeping the body at rest or uniform motion"

given

mass m= 200kg

acceleration a= 25 m/s

we know that

F= ma

F= 200*25

F= 5000 N

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Explanation:

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Mono poles cannot exist.

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If the mass of a material is 115 grams and the volume of the material is 16 cm3, what would the density of the material be?
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Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of
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Answer:

a) \dot m = 6.878\,\frac{kg}{s}, b) T = 104.3^{\textdegree}C, c) \dot S_{gen} = 11.8\,\frac{kW}{K}

Explanation:

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-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

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\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

P = 7000\,kPa

T = 500^{\textdegree}C

h = 3411.4\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

State 2s - Liquid-Vapor Mixture

P = 100\,kPa

h = 2467.32\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

x = 0.908

The isentropic efficiency is given by the following expression:

\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}

The real specific enthalpy at outlet is:

h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})

h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )

h_{2} = 2684.46\,\frac{kJ}{kg}

State 2 - Superheated Vapor

P = 100\,kPa

T = 104.3^{\textdegree}C

h = 2684.46\,\frac{kJ}{kg}

s = 7.3829\,\frac{kJ}{kg\cdot K}

The mass flow rate is:

\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}

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b) The temperature at the turbine exit is:

T = 104.3^{\textdegree}C

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0

\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})

\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )

\dot S_{gen} = 11.8\,\frac{kW}{K}

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