Question

If pressurized air pressure is 350 kPa, atmospheric pressure is 100 kPa, initial atmospheric pressure is 100 kPa, initial acceleration of the water rocket is 0.5g, acceleration of the water rocket is 0.5g, mass of water is 0.5 kg and structural mass of water is 0.5 kg and structural mass is 0.5 kg. Calculate the diameter of mass is 0.5 kg. Calculate the diameter of the nozzle where water is leaving the the nozzle where water is leaving the bottle

Answer 1

Answer:

$d=8.657mm$

Explanation:

From the question we are told that

Pressurized air pressure is $P_{air}=350 kPa,$

Atmospheric pressure is $P_a=100 kPa$

Initial acceleration of the water rocket is $a_i=0.5g.$

Acceleration of the water rocket is  $a_r=0.5g$

Mass of water is $M_w=0.5 kg$

Generally total mass is given mathematically given as

$T_M=0.5+0.5=>1kg$

Generally the tension on the rocket is given mathematically given as

$T=(P_{air}-P_a)A$

$T=(350-100) \frac{\pi d^2}{4}$

T is also

$T=\frac{3Mg}{2}$

Therefore

$T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}$

$T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}$

$d^2= \frac{3*1*9.81*4}{2(350-100) \pi}$

$d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}$

$d=8.657mm$

therefore diameter of nozzle is mathematically given as

$d=8.657mm$