Physics: Air is flowing through a rocket nozzle. inside the rocket the air has a density of 5.25 kg/m3 and a speed of 1.20 m/s. the interior diameter of the rocket is 15.0 cm. at the nozzle exit, the diameter is 2.50 cm and the density is 1.29 kg/m3. what is the speed of the air when it leaves the nozzle? air is flowing through a rocket nozzle. inside the rocket the air has a density of 5.25 kg/m3 and a speed of 1.20 m/s. the interior diameter of the rocket is 15.0 cm. at the nozzle exit, the diameter is

Air is flowing through a rocket nozzle. inside the

krojas015

09.01.2022

The velocity at exit of the nozzle is 175.8 m/s

Solution:

As per the question:

Air density inside the rocket, $\rho_{i} = 5.25\ kg/m^{3}$

Speed, v = 1.20 m/s

The inner diameter of the rocket, $d_{i} = 15.0\ cm = 0.15\ m$

The inner radius of the rocket, $r_{i} = \frac{0.15}{2} = 0.075\ m$

The exit diameter of the nozzle, $d_{e} = 2.50\ cm = 0.025\ m$

The exit radius of the nozzle, $r_{e} = 0.0125\ m$

Air density inside the nozzle, $\rho_{n} = 1.29\kg/m^{3}$

Now,

To calculate the air speed when it leaves the nozzle:

Mass rate in the interior of the rocket, $\frac{dM}{dt} = \rho_{i}Av$

Mass rate in the outlet of the nozzle, $\frac{dm}{dt} = \rho_{n}A'v'$

Now,

$A = \rho_{i}r_{i}^{2}$

Now,

We know that:

Av = A'v'

$\rho_{i}r_{i}^{2}v = \rho_{n}r_{e}^{2}v'$

$5.25\times 0.075^{2}\times 1.2 = 1.29\times 0.0125^{2}v'$

v' = 175.8 m/s

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BluSeaa

05.10.2022

$y = \frac{1}{2}x +2$

Step-by-step explanation:

From the table of function given, you would observe that if you subtract 2 from half of the x-variable values, you'd get the y-variable values.

For example, half of -8 = -4. If you add 2 to -4, you'd get: -4 + 2 = -2. Same applies to other x-values on the table.

Thus, an expression for the function represented by the table values can be written as,

$y = \frac{1}{2}x +2$

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