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
<u>t = 16.11 s</u>
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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)
<u>Pg = 1009400 Pa = 1 MPa</u>
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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
<u>P = 1110725 Pa + 1.11 MPa</u>
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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)
<u>F = 2.22 MN</u>
