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4 years ago

34. What does the D stand for in the EVADE acronym?

Engineering

1 answer:

rewona [7]4 years ago

3 0

Answer:

Even

Vicious

Animal

Die

Equally

Explanation: im jk... but its "Designate and enforce scene control zones"

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Can anyone help me ?

MrRa [10]

That’s too hard for me lol oof

3 0

3 years ago

A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature

mafiozo [28]

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

6 0

4 years ago

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$