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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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Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
sveta [45]

Answer:

V=14m/s

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut  A and bag is mathematically given as

  M_A=(60+5)*2

   M_A=130kgm/s

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

  130=(60*1)=(5*v)

   V=14m/s

7 0
1 year ago
Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
earnstyle [38]

The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.


As the coefficient of absorption would define the energy present in the reflected wave, the material C has the highest percentage of absorption i.e. 62% and would be best suitable to make a sound proof room.

4 0
2 years ago
Read 2 more answers
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

\Phi =E\times S\times COS\Theta

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
2 years ago
In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.
Stella [2.4K]

Answer:

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.

So, the workdone by the force on the cow is  

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx

= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx

= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]

= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= -209.415 J ≅ -209.42J

5 0
2 years ago
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