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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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1. Which object is farthest from the origin at t=2sec
Stolb23 [73]

Answer:

that one i know only pe not that sorry again

6 0
3 years ago
A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0
3 years ago
15. A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the
mariarad [96]

Answer:

Distance = 30m

Displacement = 6m W

Explanation:

Given the following:

Movement 1 = 18m W

Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

Therefore, Displacement of the locomotive equals:

18m W - 12m E = 6m E

3 0
3 years ago
A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the
Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

PV=nRT

P=\dfrac{WRT}{VM}

P=\dfrac{\rho RT}{M}

T=\dfrac{PM}{\rho R}

Put the value into the formula

T=\dfrac{2\times28}{1.7\times0.0821}

T=401.2\ K

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

V_{rms}=\sqrt{\dfrac{3RT}{M}}

V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}

V_{rms}=1.87\ m/s

Hence, The speed of nitrogen molecule is 1.87 m/s.

5 0
3 years ago
If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?
puteri [66]

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

\omega=\sqrt{\dfrac{k}{m}}

Put the value into the formula

\omega=\sqrt{\dfrac{13.3}{0.321}}

\omega=6.43\ rad/s

We need to calculate the velocity

Using formula of velocity

v=\omega\sqrt{A^2-x^2}

Put the value into the formula

v=6.43\times\sqrt{0.270^2-0.130^2}

v=1.52\ m/s

Hence, The speed is 1.52 m.

6 0
3 years ago
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