0.0034L
Explanation:
Given parameters:
Number of molecules of oxygen gas = 9.23 x 10¹⁹molecules
Unknown:
Volume of oxygen = ?
Solution:
Assuming the gas is at standard temperature and pressure conditions;
Volume of gas = number of moles x 22.4
we need to find the number of moles of the oxygen gas;
number of moles = ![\frac{number of particles}{6.02 x 10[tex]^{23}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20of%20particles%7D%7B6.02%20x%2010%5Btex%5D%5E%7B23%7D)
}[/tex]
Number of moles = ![\frac{9.23 x [tex]10^{19}](https://tex.z-dn.net/?f=%5Cfrac%7B9.23%20x%20%5Btex%5D10%5E%7B19%7D)
}{6.02 x 10
}[/tex]
Number of moles = 1.53 x 10⁻⁴moles
Volume of gas = 1.53 x 10⁻⁴ x 22.4 = 0.0034L
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Answer:
526.57 Pa
Explanation:
P ( pressure at the bottom of the container) = 1.049 × 10^5 pa
Using the formula of pressure in an open liquid
Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²
Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa
P( atmospheric pressure) = 1.013 × 10^5 Pa
Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g
Subtract each of the pressure from the absolute pressure at the bottom
P(bottom) - atmospheric pressure
(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa
subtract pressure due to water from the remainder
3600 - 2050.29 = 1549.71 Pa
1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g
ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa
The correct answer is Contact force:)
The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.
Explanation:
Mass of the ball, m = 0.058 kg
Initial speed of the ball, u = 11 m/s
Final speed of the ball, v = -11 m/s (negative as it rebounds)
Time, t = 2.1 s
(a) Let F is the average force exerted on the wall. It is given by :
F = 0.607 N
(b) Area of wall, 
Let P is the average pressure on that area. It is given by :
P = 0.202 Pa
Hence, this is the required solution.
