Answer:
Explanation:
Let the thickness of the film is t and the refractive index of the material of film is n.
When light travels through a sheet of thickness t, the optical path traveled is nt.
When the path of one of slit is covered by a sheet of thickness t, the optical path becomes
x = ( n - 1) t
As the one fringe is shift, so the optical path changed by one wavelength.
i.e., x = λ
So, λ = ( n - 1) t

Answer:
39 g H2O contains 1.3 ×1024molecules H2O.
Explanation:
Answer:
The lowest possible frequency of sound is 971.4 Hz.
Explanation:
Given that,
Distance between loudspeakers = 2.00 m
Height = 5.50 m
Sound speed = 340 m/s
We need to calculate the distance
Using Pythagorean theorem




We need to calculate the path difference
Using formula of path difference

Put the value into the formula


We need to calculate the lowest possible frequency of sound
Using formula of frequency

Put the value into the formula


Hence, The lowest possible frequency of sound is 971.4 Hz.
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m