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tangare [24]
3 years ago
8

How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the s

ame size carrying the same load?
Engineering
1 answer:
pantera1 [17]3 years ago
8 0

Answer:

The volume of the extra water is 2.195 ft^{3}

Solution:

As per the question:

Mass of the canoe, m_{c} = 175 lb + w

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, m_{Kc} = 38 lb + w

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

V\rho g = mg

V = \frac{m}{\rho}                                  (1)

where

V = volume

\rho = 62.41 lb/ft^{3} = density

m = mass

Now, for the canoe,

Using eqn (1):

V_{c} = \frac{m_{c} + w}{\rho}

V_{c} = \frac{175 + w}{62.41}

Similarly, for Kevlar canoe:

V_{Kc} = \frac{38 + w}{62.41}

Now, for the excess volume:

V = V_{c} - V_{Kc}

V = \frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}

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