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Solnce55 [7]
3 years ago
15

An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is t

he volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.
Physics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

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NeX [460]

Answer:

This is false

Explanation:

This is why the answer to this question is false. If these types of objects (2 points connected) should pass through same sets of 3 points, irrespective of the order that each object was plotted, we will not get identical shapes created.

The shape that is going to be created will be dependent on the pattern/order that was involved in the plotting. If it was identical, then we will have identical shapes. But if the order during plotting is different then we will have different shapes created.

Thank you!

5 0
3 years ago
A car is travelling at 70 mph. It travels for half an hour. How far has it gone?
rusak2 [61]

Answer:

35 miles

Explanation:

Since half of 70 is 35.....

6 0
3 years ago
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bezimeni [28]

Answer:

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Explanation:

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4 0
3 years ago
A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T
mart [117]
<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

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But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

  = √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E =  4.32 × 10^-13 Joules

  r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

                            = 2.775 × 10^7 rad /sec

3 0
3 years ago
A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the
SCORPION-xisa [38]
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
(4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg 
force of gravity = (mass) x (acceleration of gravity) = (360 kg) x (9.8 m/s²) = (360 x 9.8)  kg-m<span>/s² </span><span>=   </span>3,528 newtons .  
5 0
4 years ago
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