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 2.50 cm and the density is 1.29 kg/m3. what is the speed of the air when it leaves the nozzle? 45.7 m/s 29.3 m/s 88.0 m/s 123 m/s 176 m/s

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

Solution:

As per the question:

Air density inside the rocket,

Speed, v = 1.20 m/s

The inner diameter of the rocket,

The inner radius of the rocket,

The exit diameter of the nozzle,

The exit radius of the nozzle,

Air density inside the nozzle,

Now,

To calculate the air speed when it leaves the nozzle:

Mass rate in the interior of the rocket,

Mass rate in the outlet of the nozzle,

Now,

Now,

We know that:

v' = 175.8 m/s

the answer is b

explanation:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂