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raketka [301]
3 years ago
8

An asteroid revolves around the Sun with a mean orbital radius twice that of Earth’s. Predict the period of the asteroid in Eart

h years.
help
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

8 years

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between square of the orbital period and the cube of the orbital radius of any object revolving around the Sun is constant:

\frac{T_e^2}{r_e^3}=\frac{T_a^2}{r_a^3}

where:

T_e = 1 y is the orbital period of Earth

r_e is the orbital radius of Earth

r_a=2 r_e is the orbital radius of the asteroid (which is twice that of Earth)

T_a = ? is the orbital period of the asteroid

Substituting and re-arranging the equation, we find

T_a^2 = \frac{r_a^3}{r_e^3} T_e^2 = \frac{(2 r_e)^2}{r_e^3} T_e^2=8 \frac{r_e^3}{r_e^3} T_e^2=8 T_e^2 = 8 (1y)^2=8 y

so, the orbital period of the asteroid will be 8 years.

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Answer:

Esto sucede principalmente por el movimiento rotacional de las ruedas del carrito.

Como sabemos, existen dos tipos de coeficientes de fricción, el estacionario y el cinético.

En principio, la tapa se mantendrá quieta por que no tendrá la suficiente fuerza como para vencer al coeficiente de fricción estático, por lo que no podrá moverse hasta que esta reciba un pequeño impulso, como puede ser una pequeña corriente de aire. (asumo que la superficie curva de la tapita es la que esta en contacto con la rampa)

(Voy a simplificar el movimiento rotacional, pero creo que es suficiente para explicar la situación)

Ahora, en el carrito, las ruedas están conectadas a un eje.

La fuerza gravitatoria podemos pensarla que esta arriba de este eje, mas o menos en el centro del carrito, y esta fuerza intentara que el carrito baje de la rampa (lo mismo pasa para la tapita), pero a diferencia de la tapita, esta fuerza es aplicada a una distancia dada del eje (o del centro de las rueditas) lo que podemos pensar que funciona como una palanca, amplificando así la fuerza, por lo que esta vez el rozamiento estático podrá detener la superficie de la rueda que esta en contacto con el suelo, pero cuanto el centro de masa se mueva un poco, el punto de contacto entre la rueda y el suelo se moverá, de esta manera aparece lo que llamamos movimiento rotacional (si no hubiera fricción, las ruedas simplemente deslizarían sobre la rampa) que le permite al carrito ignorar en parte al coeficiente de fricción estático.

Otro factor importante, es que las tapitas suelen tener una superficie rugosa y un centro de masa desfazado, lo que no las hace buenas rotadoras.

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2 years ago
When compared to other types of waves, electromagnetic waves differ because they are
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They differ because they are transverse wave. That is their direction of travel is perpendicular to its vibrations.
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A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T
NeX [460]

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Explanation:

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2 years ago
A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
7 0
3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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