Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, 
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

where
a = width of the slit


a = 0.000167 m

a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
A region around a charged partical or object. Let me know if this works. Hope I could help you.
The air resistance on the feather would be the correct answer.
Answer:
(a) 152.85 Nm
(b) 1528.5 Nm
Explanation:
According to the formula of power
P = τ ω
ω = 2 π f
(a) f = 2500 rpm = 2500 / 60 = 41.67 rps
So, 40 x 1000 = τ x 2 x 3.14 x 41.67
τ = 152.85 Nm
(b) f = 250 rpm = 250 / 60 = 4.167 rps
So, 40 x 1000 = τ x 2 x 3.14 x 4.167
τ = 1528.5 Nm