If a bus starts a zero and then travels 10 miles north but then travels south 3.75 towards the starting point of zero, the displacement will be 10 - 3.75 = 7.5 miles
Answer:
Explanation:
so filling in:
so to 2 sig figs (which is actually not accurate, but oh well...)
KE = 450000 J
If we want to find out how high it will have to travel up a hill so that its PE is the same as the KE at this speed, we set the value for KE = to PE:
450000 = (1000)(9.8)h so

(a) 25lx
(b) 11.11lx
<u>Explanation:</u>
Illuminance is inversely proportional to the square of the distance.
So,

where, k is a constant
So,
(a)
If I = 100lx and r₂ = 2r Then,

Dividing both the equation we get

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx
(b)
If I = 100lx and r₂ = 3r Then,

Dividing equation 1 and 3 we get

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx
Lets use Dimensional analysis.

<h3>LHS:-</h3>
![\\ \sf\longmapsto s=\left[M^0LT^0\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20s%3D%5Cleft%5BM%5E0LT%5E0%5Cright%5D)
<h3>RHS</h3>

![\\ \sf\longmapsto \left[M^0LT^{-1}\right]\left[M^0L^0T^1\right]+\dfrac{1}{2}\left[M^0LT^{-2}\right]\left[M^0L^0T^1\right]^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5BM%5E0LT%5E%7B-1%7D%5Cright%5D%5Cleft%5BM%5E0L%5E0T%5E1%5Cright%5D%2B%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5BM%5E0LT%5E%7B-2%7D%5Cright%5D%5Cleft%5BM%5E0L%5E0T%5E1%5Cright%5D%5E2)
![\\ \sf\longmapsto L+\dfrac{1}{2}\left[LT^{-2}\right]\left[T^2\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20L%2B%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5BLT%5E%7B-2%7D%5Cright%5D%5Cleft%5BT%5E2%5Cright%5D)


![\\ \sf\longmapsto [L^1]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5BL%5E1%5D)
![\\ \sf\longmapsto \left[M^0LT^0\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5BM%5E0LT%5E0%5Cright%5D)
LHS=RHS
Hence verified