Answer: el tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy
Explanation:
2m/s
it has to be 20 charecters just ignore this your answer is up there
Explanation:
The expression is :
A =[LT], B=[L²T⁻¹], C=[LT²]
Using dimensional of A, B and C in above formula. So,
![A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}](https://tex.z-dn.net/?f=A%3DB%5EnC%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3D%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%7D%5D%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%7DT%5E%7B-n%7DL%5EmT%5E%7B2m%7D%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%2Bm%7DT%5E%7B2m-n%7D)
Comparing the powers both sides,
2n+m=1 ...(1)
2m-n=1 ...(2)
Now, solving equation (1) and (2) we get :
Hence, the correct option is (E).
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>
