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3 years ago

The question on science please answer correctly

Physics

1 answer:

umka2103 [35]3 years ago

3 0

The 5kg object 5m/s 1 m off the ground

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Two pulses are moving along a string. One pulse is

raketka [301]

Answer:

The answer is the 3rd option!

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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:

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

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

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

4 0

3 years ago

Bronze is what type of matter

lana66690 [7]

Bronze is a solid

Because it doesn't change shape

4 0

3 years ago

Select the correct equations that show that the gravitational potential energy of a 1000-kg boulder raised 6 m above ground leve

wel

Answer:

$U = mgh$

Explanation:

The gravitational potential energy of an object is defined as:

$U=mgh$

where

m is the mass of the object

g is the gravitational acceleration

h is the height of the object above a reference level (usually, the ground)

For the boulder in the problem,

m = 1000 kg

g = 10 N/kg

h = 6 m

Substituting,

$U=(1000)(10)(6)=60,000 J$

6 0

4 years ago

Some of the principal minerals commonly found in igneous rocks are

Verdich [7]

I believe that the answer is C. Hope this Helps:)))

3 0

3 years ago