- #1
Sirsh
- 267
- 10
Vt = (2mg/pA)1/2
where:
Vt = Terminal Velocity (ms-1) as velocity = s/t = m/s(seconds, not distance).
m = mass in kg.
g = gravity in ms-2.
p = fluid density in kgm-3
A = Area in m2
Show that equation is dimesionally correct.
I seem to get my answer wrong on the end product; ill show my working up too where I've got my answer incorrect.
Vt = (2mg/pA)1/2
m/s = (2(kg)(m/s2)/(kg/m3)(m2))1/2
(m/s)2 = (2(kg)(m/s2)/(kg/m3)(m2))
(m/s)2 = (2(kg)(m4)/((kg)(s2)(m2))
(m2/s2) = ((kg)(m2))/(s2)
As you can see the equation is quite simplified at this point, with the only problem being that both left and right sides do not equal, which shows that the equation is not dimensionally correct, i am unsure though how to get rid of this unit of kg in my equation though. can anyone help me out on this one, thanks alot!
where:
Vt = Terminal Velocity (ms-1) as velocity = s/t = m/s(seconds, not distance).
m = mass in kg.
g = gravity in ms-2.
p = fluid density in kgm-3
A = Area in m2
Show that equation is dimesionally correct.
I seem to get my answer wrong on the end product; ill show my working up too where I've got my answer incorrect.
Vt = (2mg/pA)1/2
m/s = (2(kg)(m/s2)/(kg/m3)(m2))1/2
(m/s)2 = (2(kg)(m/s2)/(kg/m3)(m2))
(m/s)2 = (2(kg)(m4)/((kg)(s2)(m2))
(m2/s2) = ((kg)(m2))/(s2)
As you can see the equation is quite simplified at this point, with the only problem being that both left and right sides do not equal, which shows that the equation is not dimensionally correct, i am unsure though how to get rid of this unit of kg in my equation though. can anyone help me out on this one, thanks alot!