1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Montano1993 [528]
3 years ago
11

Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t

he surface with lungs full of air, his lungs could be damaged. Explain why this is so.
Physics
1 answer:
Whitepunk [10]3 years ago
5 0

Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).

This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.

The gas in the diver's lungs will - expand - and can damage the alveoli.

You might be interested in
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
  • <em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

Ft is impulse

\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0
2 years ago
A car moving around a circular track with constant speed (Select all that apply.) has an acceleration with a direction that cann
BARSIC [14]

Answer:

Explanation:

Has an acceleration directed towards center of its path

5 0
3 years ago
Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr
Aleksandr [31]
<span>c. faster, less

is the answer
</span>
6 0
3 years ago
Read 2 more answers
Lake Preston's latitude is about 44.4 degrees north of the equator. Which global wind belt are
Serggg [28]

Answer:

Idk

Explanation:

Idk

5 0
3 years ago
Other questions:
  • How does a convex mirror form an image?
    11·2 answers
  • A particle has an acceleration of +6.24 m/s2 for 0.300 s. At the end of this time the particle's velocity is +9.81 m/s. What was
    5·2 answers
  • HHHHHEEEELLLPP
    13·2 answers
  • An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s
    14·1 answer
  • Heavy ions, such as alpha particles, lose kinetic energy as they travel through matter. Consider equation 31.1 or 31.2. Where do
    10·1 answer
  • A flat (unbanked) curve on a highway has a radius of 240.0 m m . A car rounds the curve at a speed of 26.0 m/s m/s . Part A What
    10·1 answer
  • The Earth moving round the Sun in a circular orbit is acted upon by a force, and hence work must be done on the Earth by this fo
    8·2 answers
  • A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
    7·1 answer
  • Explain how air pollution impacts each of Earth's systems.
    14·1 answer
  • A force of 1 N is equal to which combination of SI units?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!