Answer:
foam but thats what i heard from other people
Explanation:
We don't have enough information in the picture to answer that question.
The disembodied right hand could just as well grab the wire in the other direction, with the thumb pointing to the right.
Maybe if we knew WHY the hand is holding the wire in this direction, and what other electrical phenomenon may be involved, we might be able to say something about the current in the wire.
(Actually, we don't even know if it's a wire. It might be a soda straw, a coat hanger, or a pool cue.)
Answer:
Higher, Windward side, Condenses
Explanation:
The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.
Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.
Answer:
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:
Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:
Ahora se reemplaza en la ecuación de período:
La aceleración experimentada por el planeta es:
Se reemplaza en la ecuación de período:
La distancia del planeta con respecto al sol es finalmente despejada:
![R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7BG%5Ccdot%20M_%7Bsun%7D%5Ccdot%20%5Cleft%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%20%5Cright%29%5E%7B2%7D%7D)
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}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%5Cleft%286.674%5Ctimes%2010%5E%7B-11%7D%5C%2C%5Cfrac%7BN%5Ccdot%20m%5E%7B2%7D%7D%7Bkg%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%281.989%5Ctimes%2010%5E%7B30%7D%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cfrac%7B%2865%5C%2Ca%29%5Ccdot%20%5Cleft%28365%5C%2C%5Cfrac%7Bd%7D%7Ba%7D%20%5Cright%29%5Ccdot%20%5Cleft%2886400%5C%2C%5Cfrac%7Bs%7D%7Bd%7D%20%5Cright%29%7D%7B2%5Cpi%7D%20%5Cright%5D%5E%7B2%7D%7D)
