The answer is A ..........
Answer:
The acceleration is 2 m/s2.
Explanation:
We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:
F = m x a ---> a= F/m
a = 40 N/20 kg <em> 1N= 1 kg x m/s2</em>
a= 40 kgx m/s2/ 20 kg
<em>a= 2 m/s2</em>
To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t
= λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t
= 750 / 2(1.20)
t
= 750 / 2.4
t
= 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t
= λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t
= 750 / 4(1.50)
t
= 750 / 6
t
= 125 nm
Therefore, the minimum thickness be now will be 125 nm