Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :
<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>
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The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for
, we find:

So, the momentum of block B after the collision is -50 kg m/s.
Learn more about momentum:
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Temperature is a measure of how rapidly or slowly molecules move.<span />
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)

