Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?
Answer:
The force = 38.4 N
Explanation:
From coulombs law,
F = kq₁q₂/r² ............................ Equation 1
Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.
When, F = 6.4 N, r = d m.
6.4 = kq₁q₂/d²......................... Equation 1
When q₁' = 2q₁, q₂' = 3q₂, r = d cm
F = k(2q₁)(3q₂)/d²
F = 6kq₁q₂/d².......................... Equation 2
Dividing Equation 1 by equation 2
6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)
6.4/F = 1/6
F = 6.4×6
F = 38.4 N.
Thus the force = 38.4 N
Answer:
B.7.5m/s2
Explanation:
average acceleration=change invelocity/time taken
a=(60-30)/4
=7.5m/s2
The answer to the question is an orgamisn
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
![v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%2040%2Acos%2828%29%5C%5CV_%7Bx%7D%20%3D%2035.31%20%5Bkm%2Fh%5D)
In order to find the vertical component, we must use the sine function of the angle.
![V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]](https://tex.z-dn.net/?f=V_%7By%7D%3D40%2Asin%2828%29%5C%5CV_%7By%7D%20%3D%2018.77%20%5Bkm%2Fh%5D)
To find 
we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force 
which gives 
. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using 
where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.
