Question

The LR5 is the specialist submarine for under water rescue. The average density of sea
water is 1028 kg/ m3
i. Calculate the increase in pressure as LR5 descends from the surface to a depth of
700m.(g=10m/s2)

ii. On another descent , the LR5 experiences a total pressure of 41x 105 pa. The
entrance to the LR5 is through an access door which has an area of 3.1 m2.
calculate the force on the outside of the door
Can anybody tell this asap

Answer 1

Answer:

(i) 7196000 Pa.

(ii) 12710000 N

Explanation:

Lets start with the 1st part of the question:

(i)

density = 1028 kg/m³

height (depth) = 700 m

gravity (g)= 10 m/s²

Pressure in liquids = density × depth × gravity

                               = 1028 × 700 × 10

                               = 7196000

Therefore, the increase in pressure as LR5 descends from the surface is 7196000 Pa.

Now, the second part of the question:

(ii)

Pressure experienced = 41 × $10^{5}$

                                     = 4100000 Pa

Area of the access door = 3.1 m²

Since Pressure = $\frac{force}{area}$, force = pressure × area

Force in this case = 4100000 × 3.1

                              = 12710000 N (Newtons)

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Hope it helps

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