No.
You forgot to subtract the weight of the gizmo
What is ( A )?
How did you get the volume of the cone?
emergancy beacon flot.jpg
I have to fin the total upthrust in KN exerted on the emergancy beacon, when it is otally imersed in the ocean.
(Assume that the density of the water is 1026kg/m^3 and that the value for g is 9.81m/s^2
What i did (which i am not sure if its right) is;
Upthrust For
Cylindryical part of float = pgah (P is density) = (1026) x (9.81) x (A) x (1.20) A = (1.20 x 0.80) = 0.96m^2 so pgah (P is density) = (1026) x (9.81) x (0.96) x (1.20) = 115.94KN
Cone part of float = pgah (P is density) = (1026) x (9.81) x (A) x (0.90) A = (0.90 x 0.80) = 0.72m^2 so pgah (P is density) = (1026) x (9.81) x (0.72) x (0.90) = 67.12KN
Total Upthrust = 183.04KN
Is this correct??
Thanks for any help!!
No.
You forgot to subtract the weight of the gizmo
What is ( A )?
How did you get the volume of the cone?
sorry i am an ----- didn't do volme for cone or cylinder, tis this more like it??
Upthrust For
Cylindryical part of float = pgv (P is density) = (1026) x (9.81) x (V) Volume of Cylinder =(Pie)(r)^2 x (h)= (3.14 x 0.40^2 x 1.20) = 0.60m^2
so pgv(P is density) = (1026) x (9.81) x (0.60) = 60.39KN
Cone part of float = pgv (P is density) = (1026) x (9.81) x (v) Volume of cone = 1/3 x (pie)(r)^2 x (h) =(1/3 x 3.14 x 0.40^2 x 0.90) = 0.15m^2
so pgv (P is density) = (1026) x (9.81) x (0.15) = 15.09KN
Total Upthrust = 75.48KN
Last edited by Kelly Bramble; 05-07-2012 at 01:45 PM.