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Katarina [22]
3 years ago
13

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the

surface area of the filament of a 150 W bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation ap- pears as visible light.)
Physics
1 answer:
laila [671]3 years ago
8 0

Answer:

A = 2.1 × 10^(-4) m²

Explanation:

We are given;

Temperature; T = 2450 K

Emissivity; ε = 0.35

Bulb rating; H = 150 W

To calculate the surface area, we will use the formula;

H = AεσT⁴

Where σ is stephan boltzman constant with a value of 5.66 × 10^(−8) W/m²⋅K⁴

Making A the subject of the formula, we have;

A = H/εσT⁴

Plugging in the relevant values gives;

A = 150/(0.35 × 5.66 × 10^(−8) × 2450⁴)

A = 2.1 × 10^(-4) m²

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Answer:

80 amperes

Explanation:

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I = 160V / 2 Ω

I = 80 Amperes

Therefore the current in the circuit is 80 amperes :)

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A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500
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Answer:

A. 4.47 m/s

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As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

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An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
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Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

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At a second point downstream

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We need to calculate the depth

Using Bernoulli theorem

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Put the value into the formula

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x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

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Hence,  The depth of the water at this point is 0.938 m.

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