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lidiya [134]
4 years ago
11

An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600

kg . It is anchored to the sea bottom by a cable. The density of seawater is 1025 kg/m3. Part A What is the buoyant force on the chamber?
Physics
1 answer:
dmitriy555 [2]4 years ago
4 0

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

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Water enters a house at 1.5 m/s through a pipe with an inside radius of 1.0cm and at a pressure of 400,000 Pa. The water then tr
kupik [55]

Answer:

pressure of the water = 3.3 × 10^{5} pa

Explanation:

given data

velocity v1 = 1.5 m/s

pressure P = 400,000 Pa

inside radius r1 = 1.00 cm

pipe radius r2 = 0.5 cm  

h1 = 0 (datum at inlet)

h2 = 5.0 m (datum at inlet)

density of water ρ = 1000 kg/m³

to find out

pressure of the water

solution

we consider here flow speed in bathroom that is =  v2 and Pressure in bathroom is =  P2  

here we will use both continuity and Bernoulli equations

because here we have more than one unknown so that  

v1 × A1 = v2 × A2 × P1 +  ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²

now we use here first continuity equation for get v2

v2 = v1 \frac{A1}{A2}    

v2 = 1.5 \frac{\pi * 0.01^2}{\pi * 0.005^2}    

v2 = 6 m/s

and now we use here bernoulli eqution for find here p2 that is

P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )

P2 = 400000  - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )

P2 = 3.3 × 10^{5} pa

3 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r
KIM [24]

Answer:

\omega_f = 1.08 rad/s

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

L_i = L_f

I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f

here we know that

I_o = \frac{1}{2}mR^2

I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2

m_1 = 9.41 kg

r_1 = \frac{4}{5}(1.61) = 1.29 m

m_2 = 21.1 kg

Now from above equation

118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f

118.1(1.73) = 188.45\omega_f

\omega_f = 1.08 rad/s

8 0
3 years ago
Must physics always be explained by math?
cluponka [151]

Answer:

Because Physics is mainly composited of math

Explanation:

If you really think about it the only way any science works is off maths, whether that be measuring the amount of NaOH to put into a solution or determening the amount of sugar you want in a cake it all relys on maths

5 0
3 years ago
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tekilochka [14]

Answer:

c

Explanation:

this is because you move ( kinetic energy) your hands and clap the other person's hand, you will definitely hear sound right.

3 0
3 years ago
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