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2 years ago

What is harmonic motion

Physics

1 answer:

Tanzania [10]2 years ago

3 0

Answer: NNOOOOOOOOOOOOOOOOOOONONONO

Explanation: simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same. The force responsible for the motion is always directed toward the equilibrium position and is directly proportional to the distance from it. That is, F = −kx, where F is the force, x is the displacement, and k is a constant. This relation is called Hooke’s law.

A specific example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring, the other end of which is fixed in a ceiling. At the maximum displacement −x, the spring is under its greatest tension, which forces the mass upward. At the maximum displacement +x, the spring reaches its greatest compression, which forces the mass back downward again. At either position of maximum displacement, the force is greatest and is directed toward the equilibrium position, the velocity (v) of the mass is zero, its acceleration is at a maximum, and the mass changes direction. At the equilibrium position, the velocity is at its maximum and the acceleration (a) has fallen to zero. Simple harmonic motion is characterized by this changing acceleration that always is directed toward the equilibrium position and is proportional to the displacement from the equilibrium position. Furthermore, the interval of time for each complete vibration is constant and does not depend on the size of the maximum displacement. In some form, therefore, simple harmonic motion is at the heart of timekeeping.

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What happens when a mechanical wave travels through a medium? Question 1 options: a) Waves are transferred. b) Mass is transferr

ladessa [460]

D. Energy is transferred

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3 years ago

"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a

OlgaM077 [116]

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30° = √3/2 sin 30° = 1/2

We have P = mg = 3 Kg * 9.8 m/sec²

P = 29.4 Kg*m/ sec² P = 29.4 [N]

Py = P * cos 30° Py = 29.4 [N] * √3/2 ⇒ Py = 25.43 [N]

Px = P * sin 30° Px = 29.4 [N] * 1/2 ⇒ Px = 14.7 [N]

∑ F = m* a ⇒ ∑ Fy = 0 ∑ Fx = m *a

∑ Fy = Fn - Py = 0 Py = P*cos30° Py = 25.43 [N]

Fn = 25.43 [N]

Fr = μk * Fn ⇒ Fr = 0.19 * 25.43 ⇒ Fr = 4.83 [N]

Now

∑ Fx = m *a mg sin30° - Fr = m*a ⇒ Px - Fr = m*a

14.7 [N] - 4.83 [N] = 3 [kg] * a ⇒ 9.87 /3 = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance = x₀ + V₀*t ± at²/2 but x₀ and V₀ = 0

Then

d = ( 1/2 )*a*t² ⇒ 1 [m] * 2 = 3.29 [m/sec²] * t²

t = 0.78 sec

And finally

v = a*t ⇒ v = 3.29 *(.78) ⇒ v = 2.57 m /sec

5 0

3 years ago

Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol

Tju [1.3M]

Answer:

$R \approx 2.418\times 10^{9}\,km$

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

$T = \frac{2\pi}{\omega}$

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

$\omega = \sqrt{\frac{a_{r}}{R} }$

Ahora se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }$

La aceleración experimentada por el planeta es:

$a_{r} = G\cdot \frac{M_{sun}}{R^{2}}$

Se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }$

La distancia del planeta con respecto al sol es finalmente despejada:

$R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}$

$R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}$

Finalmente, se sustituyen las variables y se determina la distancia:

$R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}$

$R \approx 2.418\times 10^{12}\,m$

$R \approx 2.418\times 10^{9}\,km$

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3 years ago

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake s

WINSTONCH [101]

Answer: a) 5 x 10^5 kg/s b) 444 MW

Explanation:

Kinetic energy per unit mass Ke is

Ke = V^2 / 2

Ke = 3^2 / 2 = 4.5 J/kg = 0.0045 kJ/kg

Now potential energy per unit mass Pe is

Pe = g x z = 9.8 x 90 = 882.9 J/kg = 0.8829 kJ/kg

The total mechanical energy of the River per unit mass e = Ke + Pe = 0.0045 + 0.8829 = 0.88744 J/kg

M = P x V = 1000 x 500 = 5 x 10^5 kg/s

b) power generation potential of the entire river at the location Wmax = Emech = M x Emech = 500,000 x 0.88744 = 444,000kW = 444MW

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3 years ago

What are scientific theories?

Goryan [66]

A scientific theory is an explanation of an aspect of the natural world that can be repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results. Where possible, theories are tested under controlled conditions in an experiment.

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3 years ago