Question

A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate of 0.3 m/s2 until reaches a speed of 4 m/s and then lowers at a constant speed. The density of salt water is 1030 kg/m3. The submarine has a hatch with an area of 2 m2 located on the top of the submarine’s body.
a. How much time does it take for the submarine to move down 100 m?

b. Calculate the gauge pressure applied on the submarine at the depth of 100

m.

c. Calculate the absolute pressure applied on the submarine at the depth of

100.

d. How much force is required in order to open the hatch from the inside of

the submarine?

answer this step-by-step, please.

Answer 1

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa  

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

a.

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s   (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed  motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed  motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

t = 16.11 s

b.

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

Pg = 1009400 Pa = 1 MPa

c.

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

P = 1110725 Pa + 1.11 MPa

d.

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the  force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

F = 2.22 MN