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True [87]
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?
Physics
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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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
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Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

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