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blagie [28]
3 years ago
13

The complex power of a load is 10-10j VA. What component should be added in parallel with the load so that the new load has a un

ity power factor
Engineering
1 answer:
Korolek [52]3 years ago
7 0
Need points bdjdjdhdhd
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Increasing rake angle will decrease the tendency for Built-Up-Edge to form. a) True b) False
anygoal [31]

Answer:

a)True

Explanation:

While machining of ductile material and high feed and low cutting speed welding action take place between tool material and chip material this welding action is called built up edge. Built up edge action takes place due to high temperature.

To decrease the built up edge action

1. Increase the rake angle.

2. Increase the cutting speed

3. Decrease the feed rate

4. Use cutting fluid

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3 years ago
If the same amount of force were applied to all four balls in the picture, which would experience the greatest change in motion?
nydimaria [60]
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4 years ago
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Phân tích phương pháp gia công plasma
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4. What is the best explanation of electric current in a wiRe 100+ PTS
yKpoI14uk [10]

Answer:

The best explanation of 'electric current' in a wire can be defined as Electron flow because of 'electrical attraction and repulsion'. Explanation: Current can be defined as the flow of electrons. When the electron charges get attracted and the current flow happens.

Explanation:

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3 years ago
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Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an
Lera25 [3.4K]

Answer:

\dot Q_{out} = 13369.104\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0

The rate of heat transfer between the turbine and its surroundings is:

\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}

The specific enthalpies at inlet and outlet are, respectively:

h_{in} = 3076.41\,\frac{kJ}{kg}

h_{out} = 2675.0\,\frac{kJ}{kg}

The required output is:

\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW\dot Q_{out} = 13369.104\,kW

4 0
3 years ago
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