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tiny-mole [99]
3 years ago
6

For two variables in a direct proportion, what is the result of doubling one variable?

Physics
1 answer:
Alexandra [31]3 years ago
7 0
If two variables are in direct proportion, doubling one variable means the other should also be doubled because the two variables increase at the same rate.
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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
3 years ago
If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?
Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

From the question given above, the following data were obtained:

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

2×10¯⁶ = I × 300

Divide both side by 300

I = 2×10¯⁶ / 300

I = 6.67×10¯⁹ A

Thus, the current in the circuit is 6.67×10¯⁹ A

4 0
2 years ago
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers
Inessa [10]

Answer:

    v₁ = -0.8087 m / s

Explanation:

To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved

Initial instant. When they are united

        p₀ = 0

Final moment. After throwing the brick

        p_{f} = (m_man + m_skate) v1 + m_brick v2

the moment is preserved

        p₀ = p_{f}

        0 = (m_man + m_skate) v₁ + m_brick v₂

        v₁ = -  \frac{ m_{brick}   }{m_{man} + m_{skate}   }  v_{2}

the negative sign indicates that the two speeds are in the opposite direction

let's calculate

        v₁ = - \frac{2.5}{67 + 4.10}  23.0

        v₁ = -0.8087 m / s

6 0
3 years ago
A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
Basile [38]
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
3 0
3 years ago
Enter the chemical formula of a binary molecular compound of hydrogen and a group 7a element that can reasonably be expected to
LenKa [72]

The chemical formula of a binary molecular compound of hydrogen and a group 7a element that can reasonably be expected to be less acidic in aqueous solution than HBr is HF

Group 7a that forms binary molecular compounds of Hydrogen are  Halogens such as Iodine I, Bromine Br, Chlorine Cl and Fluorine F. Out of these four HF is the least acidic. This is calculated based on Acid dissociation constant. HF has the least value out of these four, so it is the least acidic.

A  binary molecular compound is a compound that consists of two non metals elements only.

Therefore, the chemical formula of a binary molecular compound of hydrogen and a group 7a element that can reasonably be expected to be less acidic in aqueous solution than HBr is HF.

To know more about binary molecular compounds

brainly.com/question/7960132

#SPJ4

5 0
2 years ago
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