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Marina CMI [18]
3 years ago
11

Explore how archemides principle is applied in building a ship and submarine​

Physics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

Principle Archimedes is applied in building a ship and submarine using the manipulating that buoyancy, is controlled the ballast tank system.

Explanation:

Submarine is rather had they focused on main parts of the submarine,he is complex and long process implementation,the most submarine design like submarine stability.

Submarine stability is complete and the fundamental Archimedes principle to arrive the weight of submarine is equal to buoyancy force.

Submarine into the parts and components of ballast tank the sequence in diving and surfacing,there two vital parts:-  flood parts and air vents

flood parts:- at the bottom position and allow water to enter or leave that tank.

air vents:- air vents at the top of the pressure hall,and that they submarine dive.

this time submarine is most modern system is depth is 300 to 450 meters,high pressure  air is 15 bar is tank air valve.

submarine is basic of the effective volume of all the submarine surfaced condition,submarine minus to the free water flood is equal to the fully pressure hull,submarine is the surfaced condition.

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How long would it take light from the sun to reach mercury?
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This is the third most abundant gas in the atmosphere This is from a crossword puzzle and it has 5 letters.
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3 years ago
Consider a circuit with two resistors in parallel R_1 = 10 ohm and R_2 = 5 ohm.A) Determine the total resistance of the circuit.
Sophie [7]

Answer:

Explanation:

Given

R_1=10 \Omega

R_2=5 \Omega

when resistance in Parallel

\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2}

R_p=\frac{R_1R_2}{R_1+R_2}

R_p=\frac{10}{3}

Suppose V is voltage of battery

Total Current i=\frac{3V}{10}

Since Circuit is Parallel therefore Voltage across both resistor is same

V=i_1R_1=i_2R_2

and i_1+i_2=i

i_1+i_1\cdot \frac{R_1}{R_2}=i

i_1(1+\frac{10}{5})=\frac{3V}{10}

i_1=\frac{V}{10}

i_2=\frac{2V}{10}

(b) When Circuit is in series

R_s=R_1+R_2

R_s=10+5=15 \Omega

since circuit is in Series therefore current is same in both resistor

Current i=\frac{V}{15} A

Voltage drop across R_1=i\times R_1

V_1=\frac{V}{15}\times 10=\frac{2V}{3}

V_2=\frac{V}{15}\times 5=\frac{V}{3}              

8 0
3 years ago
PLEASE HELP (100)
morpeh [17]

Answer:

The work done will be W=55.27\: J

Explanation:

The work equation is given by:

W=F\cdot x

Where:

F is the force due to gravity (weight = mg)

x is the length of the ramp (3 m)

Now, the force acting here is the component of weight in the ramp direction, so it will be:

F_{x-direction}=mgsin(28)

Therefore, the work done will be:

W=mgsin(28)*3

W=4*9.81*sin(28)*3

W=55.27\: J

I hope it helps you!

3 0
3 years ago
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