Recent content by 168918791999

okay to calculate the amount of gunpowder we need in total (at 100%) i assumed that the 1.6 grams is the amount of gunpowder we needed only to fly up to 75 meters because it is only 0.6% efficient it means that if i have an amount of gunpowder (100%) only 0.6% goes into flying the rocket and...

1.6 is the mass of the gunpowder so if it is 0.6% efficient that would mean: 1.6 gr | 0.6% Mr | 100%

No, that would mean we needed: 1.63525964 / 0.6 *100 = 272.543273333 grams of gunpowder That would mean that we needed more gunpowder than the rocket itself is

However thank you for your help!

p=mg*vg 3.8=mg*2323.79 mg=0.00163525964 Kg mg=1.63525964 gr Do you still have to multiply it by 10 because a rocket is on average only 10% efficient? Because then you get 16.35 gr and that would be more realistic than 1.6 gr

Keg = Eg KEg = 0.5*Mg*Vg2 Eg = 2.7x106*mg 0.5*Vg2 * mg= 2.7x106*mg 0.5*Vg2 = 2.7x106 Vg2 = 5.4x106 Vg = 2323.79 m/s = 8365.644 km/h Would this not be too fast?

KEg=P2/2mg KEg=3.82/2mg KEg=14.44/2mg Can you say KEg=Eg? Or is KEg+KEr=Eg?

Eg=Mg*2.7x106 KEg=½*Mg*Vg

P=Mg*Vg 3.8=Mg*Vg But how to calculate any of the unknowns?

P=v/g 3.8=v/9.81 V=3.8*9.81 V=37.278 m/s This is the only equation i could found

However can you say the momentum (p) is 3.8 Ns? And that the kinetic energy is KE=0.5*mg*V2 KE=0.5*mg*38.362 Ke=735.7448*mg

KE=2,7x106*mg mg = Mass of the gunpowder P=mg*v P=mg*38

F=m*Δv 3.8=0.1*Δv Δv=38m/s2

The momentum of the gunpowder must be the same as the momentum of the rocket but it is in the opposite direction. So -3.8Ns

the energy in the gunpowder is 2.7x106 J/Kg right? So 2.7x106 J/m3