Oh I’m so sorry rip winter
Answer:
vector of zero magnitude
Explanation:
The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.
In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.
If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude
Answer:
Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".
Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>
(Remember: kinetic energy =0)
Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.
While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>
(Remember: kinetic energy=0)
Due to this process the whole energy is conserved.
As the potential energy decreases kinetic energy increases.
Answer:
,
, 
Explanation:
The cube root of the complex number can determined by the following De Moivre's Formula:
![z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D%20%3D%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D%5Ccdot%20%5Cleft%5B%5Ccos%5Cleft%28%5Cfrac%7Bx%20%2B%202%5Cpi%5Ccdot%20k%7D%7Bn%7D%20%5Cright%29%20%2B%20i%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7Bx%2B2%5Cpi%5Ccdot%20k%7D%7Bn%7D%20%5Cright%29%5Cright%5D)
Where angles are measured in radians and k represents an integer between
and
.
The magnitude of the complex number is
and the equivalent angular value is
. The set of cubic roots are, respectively:
k = 0
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B1.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B1.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)

k = 1
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B3.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B3.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)

k = 2
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B5.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B5.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)
