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3 years ago

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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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Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

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$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

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