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irina [24]
3 years ago
6

If the pressure is 100,000 Pa, R is 8.314, the temperature is 300 K, and the number of moles of gas (n) is 100, what is the volu

me in m3?
Physics
1 answer:
djyliett [7]3 years ago
4 0
We can solve the problem by using the ideal gas equation:
pV=nRT
where
p is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the gas constant
T is the absolute temperature of the gas

For the gas in our problem, we have:
p=100000 Pa
R=8.314 J mol^{-1} K^{-1}
T=300 K
n=100

If we rearrange the equation and we put these numbers into it, we find the volume of the gas:
V= \frac{nRT}{p}= \frac{(100)(8.314)(300)}{100000}=2.49 m^3
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A uniform steel rod has mass 0.300 kg and length 40.0 cmand is horizontal. A uniform sphere with radius 8.00 cm and mass 0.700 k
Brrunno [24]

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1.86 cm

Explanation:

The center of gravity of the combined object x = m₁x₁ + m₂x₂ + m₃x₃/m₁ + m₂ + m₃ where m₁ = mass of sphere on left end of rod = 0.700 kg, x₁ = distance of center of mass of sphere from left end of combined object = radius of sphere = 8.00 cm, m₂ = mass of rod = 0.300 kg, x₂ = distance of center of mass of rod from left end of combined object = diameter of sphere on left end + length of rod/2 = 16.00 cm + 40 cm/2 = 16.00 cm + 20 cm = 36 cm, m₃ = mass of sphere on right end of rod = 0.580 kg, x₃ = distance of center of mass of sphere from left end of combined object = length of rod + diameter of first sphere + radius of sphere = 40 cm + 16 cm + 6.00 cm = 62 cm

x = (m₁x₁ + m₂x₂ + m₃x₃)/m₁ + m₂ + m₃

Substituting the values of the variables into the equation, we have

x = 0.700 kg × 8.00 cm + 0.300 kg × 36 cm + 0.580 kg × 62 cm/(0.700 kg + 0.300 kg + 0.580 kg)

x = 5.6 kg cm + 10.8 kg cm + 35.96 kg cm/1.580 kg

x = 52.36 kgcm/1.580 kg

x = 33.14 cm

Since the center of the rod is at x' = (40 cm + 16.00 cm + 12.00 cm)/2 = 70.00 cm/2 = 35 cm

The distance between the center of the rod and the center of gravity is x' - x = 35 cm - 33.14 cm = 1.86 cm

So, the center of gravity is 1.86 cm away from the center of the rod and closer to the 0.700 kg sphere.

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