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dedylja [7]
3 years ago
15

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

ic and what fraction potential when the displacement is one third the amplitude
Physics
1 answer:
expeople1 [14]3 years ago
7 0

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

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Answer:

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Explanation:

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let's locate the origin on the left speaker

let's find the wavelength with the equation

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we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

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we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

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therefore we see that there is respect even where the intensity is minimal

6 0
2 years ago
Assume patmos=1.00atm. what is the gas pressure pgas? express your answer in pascals to three significant figures.
hodyreva [135]
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What is the acceleration of a 2000 kg truck that has an engine exert 400 N of<br> force?
amid [387]

The acceleration of a 2000 kg truck that has an engine exerts 400 N of force <u>at 0.2 m/s².</u>

<u />

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Calculation:-

mass = 2000 kg

Force = 400 N

acceleration = F/mass

                      =  400/2000

                      = <u>0.2 m/s²</u>

<u />

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

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6 0
1 year ago
The boy soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in thi
Andrei [34K]

Answer:

(a) W(s) = M*g

(b) 7.84kg

(c) - 10.07m/s

Explanation:

Parameters given:

Mass of boy, m = 79kg

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Final velocity of boy after jumping on the skateboard, v(b) = 1.41m/s

Mass of skateboard, M = ?

Initial velocity of skateboard = u(s)

Final velocity of skateboard = v(s)

(a) The weight of the skateboard is given as the mass of the skateboard multiplied by acceleration due to gravity. Mathematically,

W = M * g

Where g = acceleration due to gravity

(b) Using the law of conservation of momentum, we have that:

m*u(b) + M*u(s) = m*v(b) + m*v(s)

We know that after jumping on the skateboard, the boy and the skateboard have the same final velocity, hence,

v(b) = v(s) = v

=> m*u(b) + M*u(s) = (m + M) * v

Making M the subject of the formula, we have that:

M = m{u(b) - v} / v

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(c) According to Newton's third law of motion, the momentum of the boy after falling must be equal to but opposite the momentum of the skateboard. This is the concept that birthed the law of conservation of momentum.

Momentum is given as:

P = m*v

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m = mass of object

v = velocity of object

Therefore, applying Newton's third law of motion:

m * v(b) = -M * v(s)

Where

v(b) is now the velocity of the boy after failing off the skateboard = 1.0m/s

v(s) is now the velocity of the skateboard after the boy falls off.

Making v(s) subject of the formula, we have that:

v(s) = m * v(b) / M

v(s) = -(79 * 1.0) / 7.8

v(s) = -10.07m/s

The negative sign means that that velocity of the boy is opposite the velocity of the skateboard.

4 0
3 years ago
An object with density of 0.8g/ml will float in water
krok68 [10]

Answer:

yes because density of object is less than water (i..e 1)

6 0
2 years ago
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