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vivado [14]
3 years ago
12

Light bulb 1 operates with a filament temperature of 2800 K, whereas light bulb 2 has a filament temperature of 1700 K. Both fil

aments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs. A1/A2
Physics
1 answer:
Veronika [31]3 years ago
5 0

Answer:

A₁/A₂ = 0.136

Explanation:

The power radiated by a filament bulb is given by the following formula:

E = σεAT⁴

where,

E = Emissive Power

σ = Stephen Boltzman Constant

ε = emissivity

A = Area

T = Absolute Temperature

Therefore, for bulb 1:

E₁ = σε₁A₁T₁⁴

And for bulb 2:

E₂ = σε₂A₂T₂⁴

Dividing both the equations:

E₁/E₂ = σε₁A₁T₁⁴/σε₂A₂T₂⁴

According to given condition, the emissive power and the emissivity is same for both the bulbs. Therefore,

E/E = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁T₁⁴/A₂T₂⁴

A₁/A₂ = (T₂/T₁)⁴

where,

T₁ = 2800 K

T₂ = 1700 K

Therefore,

A₁/A₂ = (1700 K/2800 K)⁴

<u>A₁/A₂ = 0.136</u>

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A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.50 T. The field is then adjusted so that a
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Answer:

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Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

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a=r\omega^2

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a_1=r_1\omega^2 ...... (1)

r_1=1\ m

Acceleration of Zak's is :

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Dividing equation (1) and (2) we get :

\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}

So, the acceleration of Abbie is half of the Zak's.

7 0
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