Answer:
0.73 W/m²
0.2522 W/m²
Explanation:
= Unpolarized light = 1.46 W/m²
= Analyzer angle = 54°
Light through first filter

The intensity of the light leaving the polarizer is 0.73 W/m²
After passing through analyzer

The intensity of the light that reaches the photocell is 0.2522 W/m²
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It would decrease, hope it helps
A=(v2-v1)/t
v2=0,6m/s
v1=2,4m/s
t=4s
a=(0,6-2,4)/4=-1,8/4=-0,45 m/s^2