Anyone who scuba dives is advised not to fly within the next 24 h because the air mixture for diving can introduce nitrogen into the bloodstream. Without allowing the nitrogen to come out of solution slowly, any sudden air-pressure reduction (such as during airplane ascent) can result in the nitrogen forming bubbles in the blood, creating the bends, which can be painful and even fatal. Military special operation forces are especially at risk. What is the change in pressure on such a special-op soldier who must scuba dive at a depth of 16 m in seawater one day and parachute at an altitude of 8.1 km the next day? Assume that the average air density within the altitude range is 0.87 kg/m3.

Explanation:

Let pressure at surface of earth be P Pa.  

pressure at height of 8.1 km in air can be calculated as follows .

pressure due to column of air of 8.1 km height

= h d g , h is height , d is density of air and g is acceleration due to gravity

= 8.1 x 1000 x .87 x 9.8 = 6.9 x 10⁴ Pa .

pressure at the height of 8.1 km

= P - 6.9 x 10⁴ Pa

Pressure due to  column  of 16 m in the sea

= h d g

16 x 1000 x 9.8

= 15.68 x 10⁴ Pa .

Pressure at depth of 16m

= P + 15.68 x 10⁴

pressure difference between points at height of 8.1 km and pressure at point 16 m deep

=  P + 15.68 x 10⁴ -  P +  6.9 x 10⁴ Pa

= 22.58 x 10⁴ Pa .

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.