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S_A_V [24]
3 years ago
5

I need help with this

Physics
1 answer:
Vilka [71]3 years ago
3 0

The right answer is atom. Please rate brainiest.

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Means/end analysis involves:_________
polet [3.4K]

a. a gradual approximation to the final solution

Explanation:

Means/end analysis is a process that involves creation of an end goal to enable the identified means to apply.

In this techniques sub-goals are formed to eliminate the challenges faced in application of a selected operator.

It starts by identifying a predetermined goal which is followed by actions that will led to the goal.

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Mean/end analysis : brainly.com/question/1213695

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2 years ago
The force of an electric field is proportional to to electric charge? True or False
tamaranim1 [39]

Answer:True

Explanation:

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Brainliest!! Please help with number 4 and please give me the equation too!! Brainliest!!!
Thepotemich [5.8K]

Answer:

Oooo someone is writing a answer. (Also im new to this so Idk what to really do.)

Explanation:

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Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0
2 years ago
Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.76×10^6 m, then the mass of moon will be​
igomit [66]

Answer:

7.65 x 10^22kg

sorry if im wrong!

5 0
2 years ago
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