Anyone good at physics?

[EyezsoLow 10+ year member]

Prove that the upward velocity of a rocket of initial mass M0, which is propelled by fuel burning at a rate of R kg/s, is given by vy = uex * ln[M0 / M(t) ] - g * t. uex is the speed of the exhaust gas relative to the rocket and M0 is the initial mass (rocket + fuel).

Also find the initial velocity and acceleration and velocity at time = 180s.

M0 = 2.12E6 kg

Fth = 2.32E7 N (Force of thrust)

R = 4.6E3

M(t) = M0 - R * t

Fth = -R * uex

No idea what to do or where to start...

Click to read more...

 

[Varinabassing 10+ year member]

got me lol i just made it past chemistry ha if i took the class could figure it out i am sure

 

[Ohioguy 10+ year member]

i dont understand any of that at all

 

[PioneerSPL1990 10+ year member]

N = nutons. LOL

 

[mxer69 10+ year member]

Also find the initial velocity and acceleration and velocity at time = 180s.

velocity=first derivative

acceleration=second derivative

then just plug in 180s and you'll have the value. You should know how to do that since you're in Physics

I can't read the rest. Write it down, take a picture, and I'll help. I took it freshman year so its been a while

 

[hzsogood 5,000+ posts]

Prove that the upward velocity of a rocket of initial mass M0, which is propelled by fuel burning at a rate of R kg/s, is given by vy = uex * ln[M0 / M(t) ] - g * t. uex is the speed of the exhaust gas relative to the rocket and M0 is the initial mass (rocket + fuel).
Also find the initial velocity and acceleration and velocity at time = 180s.

M0 = 2.12E6 kg

Fth = 2.32E7 N (Force of thrust)

R = 4.6E3

M(t) = M0 - R * t

Fth = -R * uex

No idea what to do or where to start...

This ois a perfect example of back in the day when you say to the teacher.. When am i ever going to need this....... wtf dde you in like astrophysics or something?

 

[Johnny Law.Lulz 5,000+ posts]

Id kill myself or drop out if i had that.

 

[mxer69 10+ year member]

its regular physics..

from what I've read (I can't read your mumbo jumbo of an equation), its just the overall acceleration of the rocket that your finding (rate of burning the fuel-rate of gravity=rate of acceleration)

 

[EyezsoLow 10+ year member]

This ois a perfect example of back in the day when you say to the teacher.. When am i ever going to need this....... wtf dde you in like astrophysics or something?

It's calculus based physics.

Here is a scan of the question, not sure if it's easier to read though

 

[hzsogood 5,000+ posts]

It's calculus based physics.

Here is a scan of the question, not sure if it's easier to read though

wow.. no that clusterfvck just nearly made it worse.. I would quit college and go work at walmart if i had to learn that shit

 

[stangman67 10+ year member]

Thats quite a bit of work. I took the calc physics series last year. I'll try to mess with it a bit.

 

[mxer69 10+ year member]

ignore these dumb *****. stay in college. it gets better. once you understand it, its quite easy.

To make things easier, take off the notation so you see actual numbers rather than scientific numbers so 4.6x10^3=4600kg/s

then just plug them in. I mean, it gives you the fricken variables lol.

only thing is, I wonder if your professor is a stickler. I mean, he could be a real *** and say that for you must also calculate in the changing weight of the rocket as it takes off. Not hard to do, just adds one more step lol.

Make sure you know how to take derivatives correctly though

 

[EyezsoLow 10+ year member]

I'd send money with paypal to anyone who solved/helped me solve it.

 

[EyezsoLow 10+ year member]

ignore these dumb *****. stay in college. it gets better. once you understand it, its quite easy.
To make things easier, take off the notation so you see actual numbers rather than scientific numbers so 4.6x10^3=4600kg/s

then just plug them in. I mean, it gives you the fricken variables lol.

only thing is, I wonder if your professor is a stickler. I mean, he could be a real *** and say that for you must also calculate in the changing weight of the rocket as it takes off. Not hard to do, just adds one more step lol.

Make sure you know how to take derivatives correctly though

Not sure what you mean "as it takes off". But yes, the weight of the rocket changes during flight and so does the acceleration. And yeah finding the values is no problem but I have to derive the equation for velocity that he gives in the question, and I'm having trouble doing that.

 

[mxer69 10+ year member]

I can solve it... How much are we talking about? I can get it to you by tomorrow morning (I've got to go to bed) so you'd be looking at 8 or 9 am Eastern