Answer:
option C
Explanation:
The correct answer is option C
A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂ medium. Hence it travels 2 t distance more than the light that is reflected off n₂.
It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).
Wavelength in n₂ is = 
Hence, path length difference = 
Answer:
6010.457N
Explanation:
Centripetal acceleration = a= V²/R
At a radius of 3.6m and velocity of 16.12m/s,
Acceleration is
a = 16.12²/ 3.6 = 72.182 m/s²
Force = Mass (m) * Acceleration (a)
36 = m * 72.182
m = 36/72.182
At breaking point
Radius = 0.468 m and Velocity = 75.1 m/s
a = V²/R = 75.1²/0.468
a = 12051.3 m/s
F = Mass(m) * Acceleration (a)
F = m * 12051.3
m = F/ 12051.3
Settings the ratio of mass equal
m = m
=> 36/72.182 = F/12051.3
F = 12051.3 * 36/72.182
F = 6010.457N
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Divide CFU of Dilution. Divide the CFU of the dilution (the number of colonies you counted) by the result from step 4. For this example, you work out 46 ÷ 1/1000, which is the same as 46 x 1,000. The result is 46,000 CFU in the original sample.
