Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
Newton's<span> First </span>Law of Motion<span>: I. Every object in a state of uniform </span>motion<span> tends to remain in that state of </span>motion<span> unless an external force is applied to it. This we recognize as essentially Galileo's concept of inertia, and this is often termed simply the "</span>Law<span> of Inertia".</span>
Answer:
The magnitude of the force net is an acting object multiplied by acceleration of the object
Explanation:
According to Rydberg's formula, the wavelength of the balmer series is given by:
R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:
Here, 
is the "general" Rydberg constant, 
is electron's mass and M is the mass of the atom nucleus
For hydrogen, we have, 
:
Now, we calculate the wavelength for hydrogen:
![\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5C%5C%5Clambda%3D%5BR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.0967%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5646%2A10%5E%7B-7%7Dm%3D656.46nm)
For deuterium, we have 
:
![R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm](https://tex.z-dn.net/?f=R_D%3D%5Cfrac%7B1.09737%2A10%5E7m%5E%7B-1%7D%7D%7B%281%2B%5Cfrac%7B9.11%2A10%5E%7B-31%7Dkg%7D%7B2%2A1.67%2A10%5E%7B-27%7Dkg%7D%29%7D%5C%5CR_D%3D1.09707%2A10%5E7m%5E%7B-1%7D%5C%5C%5C%5C%5Clambda%3D%5BR_D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.09707%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5629%2A10%5E%7B-7%7D%3D656.29nm)
