Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
</span><span>Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
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Explanation:
It is given that,
Frequency of the laser light, 
Time,
(a) Let
is the wavelength of this light. It can be calculated as :



or

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :


n = 13440
Hence, this is the required solution.
As we know,

so, let's solve for charge (q) :
time = 5 minutes = 5 × 60 seconds = 300 seconds.
hence, the charge = 60 coulombs (C)
Metals are giant structures of atoms held together by metallic bonds. “Giant” implies that large but variable numbers of Atoms are involved - depending on the size of the bits of metal. most metals are close packed - that is, they fit as many items as possible into the available volume.
The answer is the less dense plate slides over the denser plate.