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3 years ago

5

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 floa

ting 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?

1 answer:

tresset_1 [31]3 years ago

8 0

Answer:

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

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