Answer:
F'= 4F/9
Explanation:
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :
...(1)
Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :
.....(2)
Dividing equation (1) and (2), we get :
Hence, the correct option is (d) i.e. " 4F/9"
The average speed is the ratio between the total space and the total time of the motion:
The total space is
while the total time is
So, the average velocity is
We can also rewrite it in m/s. The total space is
, while the time is
, and so
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)
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Answer:
you would have to stand 6 ft back
Explanation:
