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kaziyahf2006
18.11.2020 •
Physics
The density of ice is 917 kg/m^3, and the density of sea water is 1025 kg/m^3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 6.78 m^3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?
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Ответ:
w_bear = 7175.95 N , m = 732.24 kg
Explanation:
Let's analyze the situation a bit, the ice block is in equilibrium with the thrust given by Archimedes' law that is directed towards the bottom, the weight of the ice and the weight of the bear.
B -W_ice - w_bear = 0
w_bear = B - W_ice
The thrust is given by
B = ρ g V
B = 1025 9.8 6.78
B = 68 105.1 N
Note that we use the total volume of the block since the problem indicates that it is submerged.
The weight of the ice is
W_ice = m g
the density is
ρ_ice = m_ice / V
m_ice = rho_ice V
we substitute
W_ice = ρ_ice g V
W_ice = 917 9.8 6.78
W_ice = 60929.15 N
we substitute in the first equation
w_bear = 68105.1 - 60929.15
w_bear = 7175.95 N
the mass of this bear is
w_bear = m g
m = w_bear / g
m = 7175.95 / 9.8
m = 732.24 kg
Ответ:
ok
Explanation:
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