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.
Metal ores
Explanation:
in an area where subduction has occurred in times past, metal ores are likely to be found.
Metallic ores find subduction zone regions very favorable to crystallize out of a magma.
- Ores have different modes of formation.
- Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
- These are igneous terrains where metallic sulfides and other minerals crystallize out of magmatic body.
- Metals in magma usually have large sizes and do not partition easily in the melt.
At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.
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Answer:
<em>The total potential (magnitude only) is 11045.45 V</em>
Explanation:
<u>Electric Potential
</u>
The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

The total electric potential in A is


The total potential (magnitude only) is 11045.45 V
Answer: the object transferred some of its energy to its surroundings.
Explanation: