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pishuonlain [190]
3 years ago
14

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top a

nd bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is 2π ph/g, where h is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.
Physics
1 answer:
Marysya12 [62]3 years ago
4 0

Explanation:

Equilibrium position in y direction:

W = Fb (Weight of the block is equal to buoyant force)

m*g = V*p*g

V under water = A*h

hence,

m = A*h*p

Using Newton 2nd Law

-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0

Hence, T time period

T = 2*pi*sqrt ( h / g )

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A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

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Countries with traditional economies are often less developed because:
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A person on a bike (m=90kg) is traveling 4m/s at the top of a 2m hill. What is his/her
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Answer:

Ug = 1764J

Explanation:

Ug = mgh

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The mass of Object 2 is double the mass of Object 5. The mass of Object 4 is half of the mass of Object 5 and the mass of Object
SVETLANKA909090 [29]
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
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Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
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Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.

Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
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TiliK225 [7]

Answer: A device records the time it takes sound waves to travel from the surface to the ocean floor and back again. Sound waves travel through water at a known speed. Once scientists know the travel time of the wave, they can calculate the distance to the ocean floor.

Explanation:

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