2.18 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
2.18 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
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1 Answer
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When we look out of the window of a fast-moving train, a line of sight is created (imaginary) between our eyes and the object outside. The line of sight of the distant object does not change, whereas the line of sight of the objects closer to the train changes rapidly as the train moves. So the nearby objects appear to move rapidly in the opposite direction, but the distant object remained stationary.
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According to question, we can write
Total moles of gas = n = nOxygen + nOxygen =
Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3
Viscosity = Pascal . Second
=
= x = 1, x + 2y = -1, -x + z = 1
y = -1, Z = 0
Viscosity = [P1A-1T0]
1 msD = 1mm
10 vsD = 9msD
1vsD = 0.9 MsD
L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm
Zero error = 4LC = 0.4 mm
Reading = MSR + VSR + correction
= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm
= 4.12 cm = 412 * 10-2 cm
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