a porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5m above the ground. calculate the work done by him on the luggage.....
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Answers
2.1 MEASURING ATMOSPHERIC PRESSURE
The atmospheric pressure is the weight exerted by the overhead atmosphere on a unit area of surface. It can be measured with a mercury barometer, consisting of a long glass tube full of mercury inverted over a pool of mercury:
Figure 2-1 Mercury barometer
When the tube is inverted over the pool, mercury flows out of the tube, creating a vacuum in the head space, and stabilizes at an equilibrium height h over the surface of the pool. This equilibrium requires that the pressure exerted on the mercury at two points on the horizontal surface of the pool, A (inside the tube) and B (outside the tube), be equal. The pressure PA at point A is that of the mercury column overhead, while the pressure PB at point B is that of the atmosphere overhead. We obtain PA from measurement of h:
(2.1)
where rHg = 13.6 g cm-3 is the density of mercury and g = 9.8 m s-2 is the acceleration of gravity. The mean value of h measured at sea level is 76.0 cm, and the corresponding atmospheric pressure is 1.013x105 kg m-1 s-2 in SI units. The SI pressure unit is called the Pascal (Pa); 1 Pa = 1 kg m-1 s-2. Customary pressure units are the atmosphere (atm) (1 atm = 1.013x105 Pa), the bar (b) (1 b = 1x105 Pa), the millibar (mb) (1 mb = 100 Pa), and the torr (1 torr = 1 mm Hg = 134 Pa). The use of millibars is slowly giving way to the equivalent SI unit of hectoPascals (hPa). The mean atmospheric pressure at sea level is given equivalently as P = 1.013x105 Pa = 1013 hPa = 1013 mb = 1 atm = 760 torr.
2.2 MASS OF THE ATMOSPHERE
The global mean pressure at the surface of the Earth is PS = 984 hPa, slightly less than the mean sea-level pressure because of the elevation of land. We deduce the total mass of the atmosphere ma:
(2.2)
where R = 6400 km is the radius of the Earth. The total number of moles of air in the atmosphere is Na = ma/Ma = 1.8x1020 moles.
Explanation:
Answer:
The work done by porter on the luggage is 225 J
Explanation:
Given,
A porter lifts a luggage of mass = 15 kg from the ground.
It's on his head 1.5 m above the ground.
To find: The work done by him on the luggage
Solution:
here, Force is due to the weight of the luggage.
i.e F = mg , where m = mass of luggage and g = accerleration due to gravity
or, F = 15 * 10 = 150 N
Work done can be given as:
W = F * s
where, F = force and s = displacement i.e 1.5 m
W = 150 * 1.5 = 225 J
Therefore, The work done by porter on the luggage is 225 J.