A bathysphere used for deep-sea exploration has a radius of 1.52 m and a mass of

Question

A bathysphere used for deep-sea exploration has a radius of 1.52 m and a mass of 1.16 104 kg. To dive, this submarine takes on mass in the form of sea water. Determine the amount of mass that the submarine must take on if it is to descend at a constant speed of 1.40 m/s, when the resistive force on it is 1101 N in the upward direction. The density of seawater is 1.03 103 kg/m3.

Also, A horizontal pipe 20.00 cm in diameter has a smooth reduction to a pipe 10.00 cm in diameter. If the pressure of the water in the larger pipe is 7.50 104 Pa and the pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes?

Explanation / Answer

The net force on the ship is m*g -BF - 1101N...This must produce an acceleration of 1.40m/s^2.. Let m be the mass of water it must take on..

So (1.16x10^4 +m)*9.8 - rho*g*V - 1101 =(1.16x10^4+m)*1.40

So 113680+ 9.8m - 1030*9.8*4/3**1.52^3 -1101 = 16240 +1.40m
So 113680 + 9.8m - 148409 -1101 = 16240 + 1.4m

Now (9.8 - 1.4)*m = 16240 - 113680 + 148409 + 1101 = 52070
So m = 52070/(8.4) = 6199kg

P1= 75000 Pa
P2= 60000 Pa
Diameter1 = 20 cm => Radius= 10 cm = 0.10 m
Diameter2 = 10 cm => Radius= 5 cm = 0.05 m

>> Area1 x V1 = Area2 x V2

R1^2 x x V1 = R2^2 x x V2
(0.10)^2 x V1= (0.05)^2 x V2
0.01 V1= 0.0025 V2
V1= 0.25 V2------------------ (1)

Bernoulli's equation:
>> P1 - P2 = 0.5 x Water density [V2^2 - V1^1]

By using what we got from------(1)
(75000) - (60000) = 500[ V2^2 - (0.25 V2)^2]
15000 = 500 V2^2 - 31.25 V2^2
15000 = 468.75 V2^2
V2^2 = 32
V2= 5.657 m/s

>> Flow = Area2 x V2
Flow = 0.005 x 5.657 = 0.028 m^3/s

>> Mass= Flow x Water density
Mass= 0.028 x 1000
Mass= 28 Kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.