Explanation:
C is correct.
Newton second law states that force is directly proportional to acceleration with m being the constant of variation.
![f _{net} = m(a)](https://tex.z-dn.net/?f=f%20_%7Bnet%7D%20%3D%20m%28a%29)
So
![250 = 50(a)](https://tex.z-dn.net/?f=250%20%3D%2050%28a%29)
![a = 5](https://tex.z-dn.net/?f=a%20%3D%205)
A is wrong, the constant g only happens in free fall or in vertical direction
B and D are wrong due to the mathematical error or equation error
Answer : ![F = 3.5\times10^{3}\ N](https://tex.z-dn.net/?f=F%20%3D%203.5%5Ctimes10%5E%7B3%7D%5C%20N)
Explanation :
Given that
Radius of sphere ![r = 5.90\times 10^{-15}\ m](https://tex.z-dn.net/?f=r%20%3D%205.90%5Ctimes%2010%5E%7B-15%7D%5C%20m)
The distance between the centers of the two spheres is
![r = 2\times 5.90\times 10^{-15}\ m](https://tex.z-dn.net/?f=r%20%3D%202%5Ctimes%205.90%5Ctimes%2010%5E%7B-15%7D%5C%20m)
The charge of the sphere ![q = 46\times1.6\times10^{-19} C](https://tex.z-dn.net/?f=q%20%3D%2046%5Ctimes1.6%5Ctimes10%5E%7B-19%7D%20C)
The magnitude of the repulsive force between the charges pushing them a part is
Using coulomb law
![F = \dfrac {kq_{1}q_{2}}{r^{2}}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%20%7Bkq_%7B1%7Dq_%7B2%7D%7D%7Br%5E%7B2%7D%7D)
![F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes%20%2846%5Ctimes1.6%5Ctimes10%5E%7B-19%7D%29%5E%7B2%7DC%5E%7B2%7D%7D%7B2%5Ctimes%285.90%5Ctimes10%5E%7B-15%7D%29%5E%7B2%7D%5C%20m%5E%7B2%7D%7D)
![F = 3501.3\ N](https://tex.z-dn.net/?f=F%20%3D%203501.3%5C%20N)
![F = 3.5\times10^{3}\ N](https://tex.z-dn.net/?f=F%20%3D%203.5%5Ctimes10%5E%7B3%7D%5C%20N)
Hence, this is the required solution.
W=mgh
W=(6)(9.8)(4)
W= 235.2J
Answer:
you absolute buffoon Use Ohms' Law: V = RI
V = (1x10^3)(5x10^-3) = 5 volts
Yes, this is in the range of normal household voltages.
Explanation: