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True [87]
2 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]2 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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An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
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Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

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Plugging in the values in the above equation where for a unit length L = 1 m, we have

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From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

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= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

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t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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