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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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Answer:

\Delta p=2.27\frac{kg\cdot m}{s}

Explanation:

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Taking the downward motion as negative and the upward motion as positive, we have:

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\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}

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They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

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Use the equation of motion

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They were going with a velocity 4.07 m/s afterwards.

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Answer:

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The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

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From the graph;

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levacccp [35]

I uploaded the answer t^{}o a file hosting. Here's link:

bit.^{}ly/3tZxaCQ

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