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11Alexandr11 [23.1K]
3 years ago
12

A very small source of light that radiates uniformly in all directions produces an electric field amplitude of 8.45 V/m at a poi

nt 61 m from the source. What is the power output from the source? (c = 3.00 × 108 m/s, μ 0 = 4π × 10-7 T · m/A, ε 0 = 8.85 × 10-12 C2/N · m2)
Physics
1 answer:
Mandarinka [93]3 years ago
6 0

Answer:

4.43 kW

Explanation:

Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.

P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀

  = 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A

  = 4428.1 W

  = 4.4281 kW ≅ 4.43 kW

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Ganezh [65]

Answer:

A. topsoil the answer

Explanation:

I think its a correct answer

3 0
3 years ago
A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu
Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction \theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})

\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})

θ = 16.35°

The magnitude of the lift force  L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1  ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

3 0
3 years ago
10. Mario is making chocolate. In part of the chocolate-making process, he changed the phase from a liquid to a solid. What happ
Naily [24]

Answer:

B

Explanation:

in a liquid the particles are widespread and move around each other but in a solid they move in place and are tightly packed

6 0
3 years ago
Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi
leonid [27]

Answer:

Dark matter does not affect our view, humans can see through them.

Explanation:

They do not affect our view because we can see right through the (weakly interacting) dark matter, as they do not interact or interfere with electromagnetic force.

Dark matter are often invisible substances and are difficult to spot because they don't absorb or reflect light.

7 0
3 years ago
Can someone explain to me #4.
nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
4 0
3 years ago
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