Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 

Now we know that 
So 


Answer and Explanation:
1. Evaluate the function x(t) at t=0.5

2. The period of motion T can be calculated as:

Where:

So:

3. The angular frequency can be expressed as:

Solving for k:

4. Find the derivate of x(t):

Now, the sine function reach its maximum value at π/2 so:

Solving for t:

Evaluating v(t) for 0.6603981634:

So the maximum speed of the block is:
In the negative direction of x-axis
5. The force is given by:

The cosine function reach its maximum value at 2π so:

Solving for t:

Evaluating x(t) for 3.016592654:

Therefore the the maximum force on the block is:

Coulomb's law:
F = k×q₁×q₂/r² where k ≈ 9.00×10⁹NC⁻²m²
Given values:
q₁ = +1.0C
q₂ = -1.0C
F = 650N
Substitute the terms in Coulomb's law with our given values. We will have to use the absolute value of q₂ to so the algebra works out. Solve for r:
650 = 9.00×10⁹×1.0×1.0/r²
r = 3721m
Taking significant figures into account:
<h3>r = 3700m</h3>
Potassium is the 19th element so it is B