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4 years ago

A car drives at steady speed around a perfectly circular track.

Physics

1 answer:

gayaneshka [121]4 years ago

6 0

Answer:

e. Both the acceleration and net force on the car point inward.

Explanation:

If no net force acts on the car, the car must drive in a straight line, at constant speed.

As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.

In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.

The net force that causes this acceleration, aims inward, and is called the centripetal force.

It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.

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In the electron dot symbol shown, what kind of bond connects the two nitrogen atoms in this molecule?

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triple covalent bond

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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