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Allisa [31]
3 years ago
10

Consider a 150-w incandescent lamp. the filament of the lamp is 5-cm long and has a diameter of 0.5 mm. the diameter of the glas

s bulb of the lamp is 8 cm. determine the heat flux, in w/m2, (a) on the surface of the filament and (b) on the surface of the glass bulb, and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is $0.08/kwh.
Physics
1 answer:
Allisa [31]3 years ago
8 0
<span>(a).the heat trasfer surface area and heat flux on the surface of filament are Area of Surface= µDL=3.14(0.05cm)(5cm)= 0.785 cm square qs=Q/Area of surface= 150W/0.785= 191W/cmsq.=1.91x10³x10³W/Msq (b). the heat surface on the surface of heat bulb Area of surface = 3.14xD²= 3.14(8CM)²= 201.1cm² qs=Q/Area of surface=150w/201.1cm²=0.75 w/cm²= 7500w/m² the amount and cost of electrical energy consumed during one period is Electrical Consumption=QΛt=(0.15 KW)(365X8h/yr)=438 k Wh/yr Annual cost= 438 kWh/yr)($.08/kWh)= $ 35.04 /yr</span>
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Answer:

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Explanation:

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2 years ago
Light will not pass through a pair of polaroids when their axes are
Advocard [28]

Answer:

perpendicular

Explanation:

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8 0
3 years ago
A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in
timama [110]

Answer:

KE2/KE1=0.71

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

V2 = \frac{m1}{m1+m2}*V1

The kinetic energies are:

KE1=1/2*m1*V1^2

KE2 = 1/2*(m1+m2)*V2^2

KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2

Simplifying:

KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2

The ratio will be:

KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}

KE2/KE1=0.71

5 0
3 years ago
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3 years ago
Given w= 4+,- 0.02 A=2.0+,-0.2 and 3.0+,-0.6 what is the value of w\ A y square
In-s [12.5K]

Explanation:

w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty.

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2 years ago
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