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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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A large lightning bolt is observed to have a 19500 A current and move 36 C of charge. What was its duration?
Lynna [10]

Answer:

Its duration is 1.85*10⁻³ s or 1.85 ms

Explanation:

The intensity of electric current I is defined as the amount of electric charge Q (measured in Coulombs) that passes through a section of a conductor in each unit of time. The letter I is used to name the Intensity and its unit is the Ampere (A).

The intensity of electric current is expressed as:

I=\frac{Q}{t}

where:

I: Intensity expressed in Amps (A)

Q: Electric charge expressed in Coulombs (C)

t: Time expressed in seconds (s)

Being:

  • I= 19500 A
  • Q=36 C
  • t=?

Replacing:

19500 A=\frac{36 C}{t}

Solving:

19500 A*t= 36 C

t=\frac{36 C}{19500 A}

t= 1.85*10⁻³ s= 1.85 ms (being 1 s= 1,000 ms)

<u><em>Its duration is 1.85*10⁻³ s or 1.85 ms</em></u>

8 0
2 years ago
What total mass must be converted into energy
Eduardwww [97]

This question apparently wants you to get comfortable
with  E = m c² .  But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
                                                    E = m c²

                           1.03 x 10⁻¹³ joule  =  (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

                         Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

                                   =  (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

                                   =        1.144 x 10⁻³⁰  kg .    (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature.  The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:
 
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

             =  (2,242 x 10⁶ x 86,400 x 365) joules

             =          7.0704 x 10¹⁶ joules .

How much converted mass is that ?

                                           E  =  m c²

Divide each side by  c² :    Mass  =  E / c² .
c = 3 x 10⁸ m/s

              Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

                        =        0.786 kilogram ! ! !

THAT should impress us !  If I've done the arithmetic correctly,
then roughly  (1 pound  11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>
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2 years ago
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Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?
lyudmila [28]
The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy.  That's the first choice on the list.
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3 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

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