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Paraphin [41]
3 years ago
15

A regulation basketball has a 47 cm diameter and may be approximated as a thin spherical shell.

Physics
1 answer:
Bingel [31]3 years ago
7 0

Answer:

1.1 s

Explanation:

The initial ball’s height is 3.5gsin(80.6)^{\circ}=3.453003g m\approx 3.45g m

From the law of conservation of energy, the potential energy converts to kinetic energy

PE=mgh

KE= 0.5mv^{2}

Where m is mass, g is acceleration due to gravity, h is height, v is the velocity

Here, the potential energy converts to potential energy and rotation energy

Moment of inertia of the ball, I=\frac {2mr^{2}}{3} but [tex]w=\frac {v}{r} and making r the subject then substituting it back to the equation of rotation energy we obtain rotational energy as \frac {mv^{2}}{3}

3.5sin(80.6)^{\circ}=0.5mv^{2} + \frac {mv^{2}}{3}

3.5sin(80.6)^{\circ}=\frac {5v^{2}}{6}

v^{2}=\frac {21g sin(80.6)^{\circ}}{5}

From kinematics

v^{2}=2as=2\times 3.5\times a

a=\frac {v^{2}}{7}=\ frac {21g sin(80.6)^{\circ}}{5\times 7}=\ frac {21\times 9.81 sin(80.6)^{\circ}}{5\times 7}=5.806964 m/s^{2}

Since s=0.5at^{2} then t^{2}=\frac {2s}{a} and t=\sqrt {\frac {2s}{a}}=\sqrt {\frac {2\times 3.5}{5.806964 m/s^{2}}}=1.09793 s\approx 1.1 s

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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}}

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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

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4 0
4 years ago
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