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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:

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Explanation:

We have given charge on alpha particle q=3.2\times 10^{-19}C

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Potential difference V=-3.45\times 10^{-3}volt

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From energy conservation we know that

\frac{1}{2}mv^2=qV

\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}

v=1.8180\times 10^3m/sec

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3 years ago
Each of the different colors that make up white light has a different ___________________.
Verdich [7]

Answer:

Wavelength?

Explanation:

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2 years ago
A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-
wariber [46]

Answer:

Part a)

v = 12.45 m/s

Part B)

F_n = 0.05 N

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

mgH = \frac{1}{2}mv^2 + mg(2R)

so the speed at point A is given as

mg(H - 2R) = \frac{1}{2}mv^2

v = \sqrt{2g(H - 2R)}

v = \sqrt{2(9.81)(21.9 - 2\times 7)}

v = 12.45 m/s

Part B)

Now the force equation at point A is given as

F_n + mg = \frac{mv^2}{R}

F_n = \frac{mv^2}{R} - mg[/tex]

F_n = 0.004(\frac{12.45^2}{7} - 9.81)

F_n = 0.05 N

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3 years ago
If the cross-section of a wire of fixed length is doubled, how does the resistance of that wire change? (this is for studying fo
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If the cross-section of a wire of fixed length is doubled,  the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>
6 0
3 years ago
Dua titik yang bermuatan listrik yang sama, mula-mula berjarak 5 cm dan saling tarik menarik dengan gaya 225 N. Agar kedua titik
Serggg [28]

Answer:

15 cm

Explanation:

Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:

F = Kq₁q₂ / d²

Perbanyak silang

Fd² = Kq₁q₂

Menjaga Kq₁q₂ konstan, kita memiliki:

F₁d₁² = F₂d₂²

Dengan rumus di atas maka diperoleh jarak sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

F₁d₁² = F₂d₂²

225 × 5² = 25 × d₂²

225 × 25 = 25 × d₂²

5625 = 25 × d₂²

Bagilah kedua sisinya dengan 25

d₂² = 5625/25

d₂² = 225

Hitung akar kuadrat dari kedua sisi

d₂ = √225

d₂ = 15 cm

Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N

4 0
2 years ago
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