Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×
m)
d = (3.30 × 563 ×
) ÷ (0.0047)
= 1.8579 ×
÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
Answer:
resistance- decreases current-increases
The conclusion that is best supported by the data is;
D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.
Answer:
No. Because it would correspond to zero Instantaneous acceleration.
Explanation:
hope this helps
Given that,
Frequency emitted by the bat, f = 47.6 kHz
The speed off sound in air, v = 413 m/s
We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

or

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.