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3 years ago

6

Homes are often insulated with fiberglass insulation in their walls and ceiling.The thermal conductivity of fiber glass is 0.040

W/m⋅K. Suppose that the total surface area of the walls and roof of a windowless house is 370m2 and that the thickness of the insulation is 10 cm. At what rate does heat leave the house on a day when the outside temperature is 30∘C colder than the inside temperature?

1 answer:

denis-greek [22]3 years ago

3 0

Answer:

4440 W

Explanation:

k = Heat conduction coefficient = 0.04 W/(m·°C)

A = Area = 370 m²

l = Thickness = 10 cm

$\Delta T$ = Difference in temperature = 30°C

Rate of heat transfer is given by

$Q=\dfrac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\dfrac{0.04\times 370\times 30}{10\times 10^{-2}}\\\Rightarrow Q=4440\ W$

The rate of energy transfer is 4440 W

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2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

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Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

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When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
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When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

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$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

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Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

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The potential energy of the ball as it sits on top of the building is given by

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The potential energy of the ball as it is halfway through the fall is given by

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$KE=\frac{1}{2}(2)(19.8)^2=392 J$

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