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sashaice [31]
1 year ago
7

9. A balloon is filled with 1000 cm3 of a gas weighing 1000 g. Will it rise or fall when it is released?

Physics
1 answer:
torisob [31]1 year ago
3 0

We can calculate the density of the balloon as follows:

\rho=\frac{mass}{volume}=\frac{1000g}{1000cm^3}=\frac{1g}{cm^3}

Therefore, the balloon will fall

Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.

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Why is vinegar considered a solution?
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Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
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Answer:

Temperature at the exit = 267.3 C

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For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

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as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

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for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

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3 years ago
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