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pantera1 [17]
3 years ago
11

A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x

), with x in meters and F in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?
Physics
1 answer:
sattari [20]3 years ago
6 0
If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?
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A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit
Ivahew [28]

Answer: c) 90 m/s

Explanation:

Given

Invest velocity, v1 = 250 m/s

Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg

Outlet specific enthalpy, h2 = 297.31 kJ/kg = 297310 J/kg

Outlet velocity, v2 = ?

0 = Q(cv) - W(cv) + m[(h1 - h2) + 1/2(v1² - v2²) + g(z1 - z2)]

0 = Q(cv) + m[(h1 - h2) + 1/2(v1² - v2²)]

0 = [(h1 - h2) + 1/2(v1² - v2²)]

Substituting the values of the above, we get

0 = [(270110 - 297310) + 1/2 ( 250² - v²)

0 = [-27200 + 1/2 (62500 - v²)]

27200 = 1/2 (62500 - v²)

54400 = 62500 - v²

v² = 62500 - 54400

v² = 8100

v = √8100

v = 90 m/s

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