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3 years ago

8

The diagram below shows a circuit with a

1 answer:

kirill [66]3 years ago

4 0

Answer:

15v

Explanation:

V = IR

= 0.3*50

= 15V.

I HOPE IT'S HELPFUL

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Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinet

vladimir2022 [97]

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

$T_1 =$ Temperature at inlet of turbine

$T_2 =$ Temperature at exit of turbine

$P_1 =$ Pressure at exit of turbine

$P_2 =$ Pressure at exit of turbine

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

$m_i = m_0 = m$

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

Where,

m = mass

$m_i$ = mass at inlet

$m_0$ = Mass at outlet

$h_i$ = Enthalpy at inlet

$h_0$ = Enthalpy at outlet

W = Work done

Q = Heat transferred

$V_i$ = Velocity at inlet

$V_0$ = Velocity at outlet

$Z_i$ = Height at inlet

$Z_0$ = Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + W$

$W = h_i -h_0$

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

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

$\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}$

$T_2 = 725.126K$

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

So:

$W = h_i -h_0$

$W = C_p (T_1-T_2)$

$W = 1.005(1400-725.126)$

$W = 678.248kJ/Kg$

Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

5 0

4 years ago

According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

$R=\dfrac{V}{I}$

The unit of voltage is volt and that of current is ampere

Unit of resistance :

$R=\dfrac{\text{volt}}{\text{ampere}}$

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0

3 years ago

Read 2 more answers

During the first 5 seconds after a ball is dropped from rest, how far will it fall?

agasfer [191]

The answer is 125metre

5 0

3 years ago

aniked [119]

Without air resistance, the ball would reach a height of h = 31.9 m(pic 1)
The gravitational potential energy difference would be ΔU = 62.588J - 43.164J = 19,424J (pic2 + 3)
The work done by air resistance must be equal to the energy difference(pic4)
The magnitude of the air resistance: F = 0.883N

6 0

3 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

8 0

3 years ago