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2 years ago

Describa con sus palabras que fue lo que descubrio galileo en su legendario experimento en la torre inclinada de pisa

Physics

1 answer:

german2 years ago

5 0

Answer:

En 1589 Galileo realizó un experimento lanzando dos bolas de diferentes masas desde la famosa Torre Inclinada de Pisa para demostrar que el tiempo de caída es independiente de la masa de la bola. A través de este experimento, Galileo descubrió que los cuerpos caían casi simultáneamente, refutando la teoría de Aristóteles de que la tasa de caída era proporcional a la masa del cuerpo.

Debido a la imperfección de los equipo de medición de esa época, la caída libre de los cuerpos era casi imposible de estudiar. En busca de una forma de reducir la velocidad de movimiento, Galileo reemplazó la caída libre por rodar sobre una superficie inclinada, donde había velocidades y resistencia del aire significativamente más bajas. Se notó que con el tiempo, la velocidad del movimiento aumenta: los cuerpos se mueven con aceleración. Se concluyó que la velocidad y la aceleración no dependen ni de la masa ni del material de la pelota.

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A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

2 years ago

What effects can occur when heat energy is added to a system?

choli [55]

I think there would be three answers from the choices. The effects that can occur when heat energy is added to a system would be that increase in temperature, <span>change of state and phase change Hope this answers the questions. Have a nice day.</span>

6 0

3 years ago

A current of 0. 82 a flows through a light bulb. how much charge passes through the light bulb during 94 s?

gladu [14]

A current of 0. 82A flows through a light bulb. The charge passed through the light bulb during 94 s is 77.08C

The amount of charge flown for a given period of time determines the current passed through a bulb or electrical body.

The relation between the charge, current and time is given as:

Q = I × t

where, Q is the charge flown through bulb

I is the current passed through bulb

t is the time for which charge passes through bulb

Given,