gravity dwn, your hand up and equal to grav
The answer would be : <span>C. most of the blue and green has been redirected.
Hope this helps !
Photon </span>
Answer:
The magnitude of each charge is ![2.82\times10^{-6}\ C](https://tex.z-dn.net/?f=2.82%5Ctimes10%5E%7B-6%7D%5C%20C)
Explanation:
Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.
We need to calculate the magnitude of each charge
Using formula of force
![F=\dfrac{kq^2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq%5E2%7D%7Br%5E2%7D)
Where, q = charge
r = separation
Put the value into the formula
![20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}](https://tex.z-dn.net/?f=20%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes%20q%5E2%7D%7B%286%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}](https://tex.z-dn.net/?f=q%5E2%3D%5Cdfrac%7B%286%5Ctimes%2010%5E%7B-2%7D%29%5E2%5Ctimes20%7D%7B9%5Ctimes10%5E%7B9%7D%7D)
![q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}](https://tex.z-dn.net/?f=q%3D%5Csqrt%7B%5Cdfrac%7B%286%5Ctimes%2010%5E%7B-2%7D%29%5E2%5Ctimes20%7D%7B9%5Ctimes10%5E%7B9%7D%7D%7D)
![q=2.82\times10^{-6}\ C](https://tex.z-dn.net/?f=q%3D2.82%5Ctimes10%5E%7B-6%7D%5C%20C)
Hence, The magnitude of each charge is ![2.82\times10^{-6}\ C](https://tex.z-dn.net/?f=2.82%5Ctimes10%5E%7B-6%7D%5C%20C)
Answer:
Horizontal component = 241 N
Vertical component = 287 N
Explanation:
Given:
Force = F = 375 N
Referring to diagram attached, the force F is making an angle
theta = 20+30 = 50 with the horizontal.
Horizontal component = F*cos(theta) = 375*0.64278 = 241 N
Vertical component = F*sin(theta) = 375*0.76604 = 287 N
Answer:
150 steps south
Explanation:
250 north 250 back to start then continue south for remainder of 400 steps. 150 south