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Gnesinka [82]
2 years ago
5

A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the

surface (where the pressure is 1.00 atm). The temperature at the bottom is 4.0 ∘C, and the temperature at the surface is 23.0 ∘C.
a. What is the ratio of the volume of the bubble as it reaches the surface (Vs) to its volume at the bottom (Vb)?
b. Would it be safe for Jacques to hold his breath while ascending from the bottom of the lake to the surface?
Physics
1 answer:
kodGreya [7K]2 years ago
8 0

Answer:

3.73994

No,

Explanation:

P_1 = Pressure at the bottom of the lake = 3.5 atm

P_2 = Pressure at the top of the lake = 1 atm

V_b = Volume at the bottom of the lake

V_s = Volume at the top of the lake

T_1 = Temperature at the bottom of the lake = 4 °C

T_2 = Temperature at the top of the lake = 23 °C

From ideal gas law we have the relation

\dfrac{P_1V_b}{T_1}=\frac{P_2V_s}{T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{P_2T_1}{P_1T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{1\times 277.15}{3.5\times 296.15}\\\Rightarrow \dfrac{V_s}{V_b}=0.26738^{-1}\\\Rightarrow \dfrac{V_s}{V_b}=3.73994

The ratio is 3.73994

As Jacques is ascending if he holds his breath his lungs acting like a bubble would expand. Hence, it is not safe to hold his breath while ascending,

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• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

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Distance is proportional to focal length, so

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M_2=-\frac{d'_1}{d_1}

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M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

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\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

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\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

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