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3 years ago

I have been given this question

1 answer:

6 0

Frequency is no. of times this wave vibrates or complete its one cycle per unit time. As it vibrates 30000 times per second its frequency is 30000 per second or 30000 Hz [1 Hz = once per second].

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Use what you know about reflection and absorption of light to complete the sentence.

Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

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3 years ago

Read 2 more answers

Gravity is ____________________ to the distance between and difference in mass between two objects

olga55 [171]

Gravity is proportional to its mass and distance between it and another object

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3 years ago

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An object's inertia causes it to come to a rest position.

stellarik [79]

Answer:

Explanation:

the inertia of an object does not make it to come to rest, this is normally caused by friction

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2 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:

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

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

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

4 0

3 years ago

If something is hot, does it contain heat or thermal energy?

zysi [14]

I am absolutely sure that if something is hot, it contain thermal energy only because if a given body is heated its electrons start to move very fast which means that thermal energy is being produced. Hope everything is clear! Regards!

8 0

3 years ago