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Arada
3 months ago
12

A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur

faces of the pane is 15° C.
What is the rate of heat flow through this window? (Thermal conductivity for glass is 0.84 J/s⋅m⋅°C.)
a. 50 000 J/sb. 2 500 J/sc. 1 300 J/sd. 630 J/s
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
3 0

To tackle this problem, one must utilize the ideas related to the rate of heat flux defined in energetic terms. The heat transfer rate signifies the quantity of heat moved per time unit through a specific material. This can be calculated mathematically in the following manner:

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

Where

k = 0.84 J/s⋅m⋅°C (This is the thermal conductivity of the substance)

A = 1m^2 Area

L = 5*10^{-3}m Length

T_H= Temperature of the "hot" reservoir

T_C= Temperature of the "cold" reservoir

Substituting our specific values provides us with,

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

Thus, the accurate option is B.

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The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

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4 months ago
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A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer
Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

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Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

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9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

t=\frac{6\pi }{3.18}

t=5.92 s

7 0
3 months ago
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