Answer: Decreasing the distance of the space shuttle from Earth .
Explanation:
According to expression of gravitational force:
G = gravitational constant
= masses of two objects
r = Distance between the two objects.
F = Gravitational force
From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.
So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.
Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.
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)
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Answer:
The acceleration is 
Explanation:
From the question we are told that
The lift up speed is 
The distance covered for the take off run is 
Generally from kinematic equation we have that
Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest
So
=>
