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Answer:
[tex]d=8.657mm[/tex]
Explanation:
From the question we are told that
Pressurized air pressure is [tex]P_{air}=350 kPa,[/tex]
Atmospheric pressure is [tex]P_a=100 kPa[/tex]
Initial acceleration of the water rocket is [tex]a_i=0.5g.[/tex]
Acceleration of the water rocket is [tex]a_r=0.5g[/tex]
Mass of water is [tex]M_w=0.5 kg[/tex]
Generally total mass is given mathematically given as
[tex]T_M=0.5+0.5=>1kg[/tex]
Generally the tension on the rocket is given mathematically given as
[tex]T=(P_{air}-P_a)A[/tex]
[tex]T=(350-100) \frac{\pi d^2}{4}[/tex]
T is also
[tex]T=\frac{3Mg}{2}[/tex]
Therefore
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}[/tex]
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}[/tex]
[tex]d^2= \frac{3*1*9.81*4}{2(350-100) \pi}[/tex]
[tex]d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}[/tex]
[tex]d=8.657mm[/tex]
therefore diameter of nozzle is mathematically given as
[tex]d=8.657mm[/tex]