'N' is labelled 'runoff'. That's when the water is down. Streets are wet, worms come out of the ground, and water flows down the street to the sewer.
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).
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
answer is 10km
Explanation:
use "S =Ut "
S=distance U=velocity t =time
no need to convert time into seconds as the velocity has given in meters per minute