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AnnyKZ [126]
3 years ago
15

Although it shouldn’t have happened, on a dive I fail to watch my SPG and run out of air. If my buddy is close by, my best optio

n is to __________. Another option is to _____________, if I’m in shallow water and the surface is closer than my buddy.
A ascend using my buddy's alternate air source / make a controlled emergency swimming ascent




B ascend using my buddy's alternate air source / make an buoyant emergency ascent




C make a normal ascent / ascend using my buddy's alternate air source




D make a controlled emergency swimming ascent / make a normal ascent
Physics
1 answer:
Airida [17]3 years ago
5 0

Answer:

A) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent

Explanation:

When it is found that you are out of air while under water, first of all don't panic, look for your buddy. If you are unable to do that so, then you need to make an emergency ascent. First try to make a <u>Controlled Emergency Swimming Ascent</u> (CESA). This ascent remains under control and is performed at a safe ascent rate. As you ascend the air in your lungs will expand with decreasing ambient pressure. To avoid an over pressurization injury, always exhale a continuous string of bubbles while going up.

If you are not sure you will make to the surface that leading to inhale the only option is to turn the CESA into a Buoyant Emergency Ascent. To be ready locate your weight system as you ascend. As an addition remove the weight from one of weight pockets and hold it away from your body in preparation of dropping if necessary. Dropping the weight will give you an upward buoyant force which is an <u>uncontrolled buoyant emergency ascent</u> and <u><em>should be performed only as the last option</em></u>.

So, according to this, first, always have a look at your SPG. Then, if you are out of air, look for your buddy, if not found then make CESA and the last option will be buoyant emergency ascent.

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Answer:

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Now,buyantant force

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so;

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{v}^{b}  =  \frac{50}{1000 } \times 9.8

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Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

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specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

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