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Licemer1 [7]
3 years ago
7

A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end

from the mass m, the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity as the rod swings through its lowest (vertical) position?
Physics
1 answer:
konstantin123 [22]3 years ago
3 0

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

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Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.
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To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

\frac{Q}{t} = \frac{kA\Delta T}{d}

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, \Delta T, T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}

k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}

\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}

\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}

\Delta T_{air} = 44.9 \Delta T_{glass}

There are two layers of Glass and one layer of Air so the total temperature would be given as,

\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}

\Delta T = 2\Delta T_{glass} +\Delta T_{air}

20\°C = 46.9\Delta T_{glass}

\Delta T_{glass} = 0.426\°C

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\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}

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\Delta {Q}{t} = 179W

Therefore the correct answer is D. 180W.

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