Recent content by beamer09

In terms of area.

Would a 2 cm nozzle work better?

And as far as the Newtons needed to lift the bottle, I used the formula F=M*A(or G). I got a Force of 2.1805 Newtons needed to lift the bottle without water. I can not use this answer yet because I need to find what my mass would be with water. But, am i on the right track and have the right...

Well the mass of the rocket now is .2225 kg. I found the area of my nozzle which would be .2124 cm^2. Then i used the formula F= P*A to find the thrust force of my rocket which was 689.48 kpa that I am using, and the area of the nozzle which was .2124 cm^2, to get the answer of 146.4456 N. That...

One last quick question, how can I find the amount of pressure leaving the nozzle? In order to have the pressure escaping equal to the ambient pressure, I need to find out what the amount of pressure would be, in order to know if my nozzle is a sufficient one. The ambient pressure would be 100...

How do I find the efficiency of the nozzle? By the effeciency of the nozzle, what do you mean?

Now I understand your answer. Thank you very much. This will be a very big help.

Oh, wait nevermind i forgot that the question was how much force needed for lift, not how high it would go.

no, not the water that is why the answer wouldn't be one hundred percent accurate, i did not take into consideration the weight of the water. But, the water is constantly leaving the bottle and the weight will go down slowly causing it to be almost impossible to calculate a very accurate answer...

Follow up- When the water is coming out of the rocket at a force of 100 psi, is it a sufficient amount to provide lift-off?

Hello, I need to figure out the force I require to lift the water rocket that I'm building. It weighs about three ounces or about .0850 kilograms to possibly make it easier. How much force will I need to provide lift-off. Would I take the acceleration due to gravity (9.8) times the weight of the...