Recent content by danield

i am still confused as how to do this... should i use Vat nozzle divided by height and that should be the answer?

well what i did was the following, i was given the area of the fan and the volumetric flow rate, so then i assumed the nozzle will operate at steady state so A1V1=A2V2 and A1V1 is equal to the volumetric flow rate so i divided this by the area of the nozzle tip giving me the velocity at that...

we are designing a toy for a class, and now we are doing engineering models. I am supposed to do an engineering model on a fan that is supposed to levitate a ball. I was able to model the drag force generated at the tip of the nozzle, but i haven't been able to do so for different heights above...

if it is not too much to ask can you help me understand the equation? i am completely lost just from watching it.. Im assuming Umax is the velocity at the tip of the nozzle, and Function U is the velocity at certain distance away from the tip of the nozzle... I am also assuming rho is the...

Ok i have the following problem, I have a fan connected to the nozzle, i am given the nozzle area and the volumetric flow rate generated by the fan. How can i find the velocity of the air at different distances from the tip of the nozzle? any help is greatly appreciated

Homework Statement I have a mixture of saturared water liquid vapor mixture at 200 c, volume = .03 m^3 at 15.54 Bars... the mixture undergoes an isothermal expansion and the volume doubles, what is the new pressure? mtotal = .301 kg Homework Equations this is my prob not sure which...

Well this is my first diff eqs homework and I am totally lost, i have no idea what to do here are the questions that i have... Homework Statement 1)An object released from a height h meters above the gorund with a veritcal velocity of Vo m/s htis the ground after To seconds. Neglecting...

bump...

I have tried everything... but still i can't get to the answer.. i get to the formula a=(m2-m1)g/(1/2M+m1+m2) but after that i don't know how to apply the torque... :( need help

divided because to cancel out KM into M u must multiply by 1000, and to turn hours into seconds you must divide by 3600... therefore you must divide km/h by 3.6 to turn something into m/s

Homework Statement Two Masses are suspended from a pulley. The pulley itself has a mas of .20kg a radius of .15m a constant torque of .35 MN due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1= .4kg and m2= .8...

ok thanks very much

Yes i understand that part, but what i don't know is how does the up and down acceleration influence the total acceleration, this is just a hunch but i think that when acceleration is downward i would be 7.81, and when it is upward it will be 11.81, I am not totally sure about that As for the...

so that means that when the velocity is up or down 5m/s constat there is no acceleartion, so only G is taken into account? but for the toher two, how are they calculated?

Homework Statement 1) Find the period of a pendulum 50 cm long when it is suspended in (a) a stationary elevator; (b) an elevator falling at the constant speed of 5.0 m/s; (c) an elevator falling at the constant acceleration of 2.0 m/s2; (d) an elevator rising at the constant speed of 5.0 m/s...