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3 years ago

6

An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th

e top 16.0°C, what is the volume of the bubble just before it reaches the surface?

1 answer:

AlexFokin [52]3 years ago

6 0

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

so

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$ Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

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Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

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