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melomori [17]
3 years ago
6

"Two point charges exert a 6.50 N force on each other. What will the force (in N) become if the distance between them is increas

ed by a factor of 8?"
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

0.102N

Explanation:

According to Coulomb law, the charge force between 2 points charge is directly proportional to the product of the charges and inversely proportional to the square of distance between the point charges. This means if the distance between them is increased by a factor of 8, then the charge force must be decreased by a factor of 8*8 = 64.

So the new charge force F = 6.5 / 64 = 0.102 N

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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}

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

The mass flow rate is:

\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}}

\dot m = 6.878\,\frac{kg}{s}

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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3 years ago
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