Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
The maximum speed of the donkey is 10.72m/s
The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.
It is given that donkey attains an acceleration of 1.6 m/s^2
The time taken to accelerate to given speed is 6.7 seconds
We use the formula v=u + at to find the fastest speed
v is the final or maximum speed
u is the initial speed which in this case is 0 as the donkey is at rest
a is the acceleration of the donkey
t is the time taken in seconds
v = u + at
v= 0 + 1.6 x 6.7
= 10.72 m/s
Hence the donkey obtains the speed of 10.72 m/s
For further reference:
brainly.com/question/24478168?referrer=searchResults
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The person does not sink into the snow because the force acts on a larger area so that the pressure is less
The edge of the sharp knife has a smaller area so the force acting on the knife produces a larger pressure
ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
█ Question <span>█
</span><span>In an electronic transition, an atom cannot emit what?
</span>█ Answer █
When an electronic transition is occurring, an atom cannot emit ultra-violet light.
<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>
