Question

A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume in cubic meters of the submerged part of her body?

Answer 1

Any ideas?

Answer 2

it is a short response question.

Answer 3

If I call with V the requested volume, then I can write:
\[\large V \times {\rho _{{H_2}O}} \times g = P\]
where:
\[\large g = 9.81\;m/{\sec ^2}\]
\[\large {\rho _{{H_2}O}}\] is the density of water,
and P=610 N

Answer 4

and the density of water would be 999.97 kg/m^3, if that has any significance here...

Answer 5

approximately 1,000 Kg/m^3

Answer 6

yes. so v x 999.97 x 9.81 = 610?

Answer 7

that's right!

Answer 8

16.1?

Answer 9

I got 0.0621 m^3

Answer 10

whoops, divided the wrong side!

Answer 11

can you help me with one more?

Answer 12

yes!

Answer 13

Huckleberry Finn's wooden raft measuring 4.9 m by 3.7 m by 0.36 m had a mass of 977 kg. If the raft is ti stay afloat in fresh water, what is the maximum amount of extra mass in kilogram that can be safely added?

Answer 14

@Michele_Laino

Answer 15

hint:
here we can write the subsequent equation (Archimede's principle):
\[\large \left( {{m_{add}} + {m_{raft}}} \right) \times g = V \times {\rho _{{H_2}O}} \times g\]
where:
\[{{m_{add}}}\] is the added mass

Answer 16

and:
\[{{m_{raft}}}\] is the mass of the raft

Answer 17

\[(m _{add} + 977) \times 9.81 = 6.5268 \times 999.97 \times 9.81\]

Answer 18

and:
\[\large V = 4.9 \times 3.7 \times 0.36 = ...?\]

Answer 19

that's right!

Answer 20

\[m_{add} = 5549.6\]

Answer 21

that's right!

Answer 22

Thank you!

Answer 23

Thank you!