Answer:
The volume of the extra water is [tex]2.195 ft^{3}[/tex]
Solution:
As per the question:
Mass of the canoe, [tex]m_{c} = 175 lb + w[/tex]
Height of the canoe, h = 21.5 ft
Mass of the kevlar canoe, [tex]m_{Kc} = 38 lb + w[/tex]
Now, we know that, bouyant force equals the weight of the fluid displaced:
Now,
[tex]V\rho g = mg[/tex]
[tex]V = \frac{m}{\rho}[/tex] (1)
where
V = volume
[tex]\rho = 62.41 lb/ft^{3}[/tex] = density
m = mass
Now, for the canoe,
Using eqn (1):
[tex]V_{c} = \frac{m_{c} + w}{\rho}[/tex]
[tex]V_{c} = \frac{175 + w}{62.41}[/tex]
Similarly, for Kevlar canoe:
[tex]V_{Kc} = \frac{38 + w}{62.41}[/tex]
Now, for the excess volume:
V = [tex]V_{c} - V_{Kc}[/tex]
V = [tex]\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}[/tex]
