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slava [35]
3 years ago
8

A bubble of air is rising up through the ocean. When it is at a depth of 20.0 m below the surface, where the temperature is 5.00

°C, its volume is 0.80 cm3. What is the bubble's volume (in cm3) just before it hits the ocean surface, where the temperature is 20.0°C? Assume the average density of sea water is 1,025 kg/m3. Hint: Use Pascal's Principle (textbook Eq. 14.4) to determine the pressure at the depth of 20.0 m below the surface.
Physics
1 answer:
kotegsom [21]3 years ago
3 0

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

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Answer:

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4 0
2 years ago
Determine the work done by the constant force. The locomotive of a freight train pulls its cars with a constant force of 15 tons
Digiron [165]

Answer:

5.92×10⁷ J

Explanation:

We'll begin by converting 15 tons to Newton. This can be obtained as follow:

1 ton = 9806.65 N

Therefore,

15 ton = 15 ton × 9806.65 N / 1 ton

15 ton = 147099.75 N

Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:

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0.25 mi = 0.25 mi × 160934 / 100 mi

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Wd = 147099.75 × 402.335

Wd = 5.92×10⁷ J

Thus, the Workdone is 5.92×10⁷ J

5 0
2 years ago
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igor_vitrenko [27]

The formula for half-life is:

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A_{final}=20(\frac{1}{2})^{\frac{40}{8}=0.625g

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.

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