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Alecsey [184]
3 years ago
7

The intensity at a distance of 6 m from a source that is radiating equally in all directions is 6.0\times 10^{-10} W/m^2. What i

s the power emitted by the source
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

P = 271 nW

Explanation:

  • If the source is radiating equally in all directions, it can be treated as a point source, so all points located at the same distance of the source, have the same intensity I, which is related to the power by the following expression:

       I = \frac{P}{A} =\frac{P}{4*\pi *r^{2} } (1)

  • Solving for P, we get:

       P = I*4*\pi *r^{2} = 6.0e-10 W/m2 * 4 * \pi *(6m)^{2} =271 nW (2)

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Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it 
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In what way would a digital thermometer be preferable to those from a liquid-based thermometer?
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5 0
3 years ago
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
3 years ago
How deep is the outer core beneath the surface
sukhopar [10]
Precisely around 1,800 miles below.
6 0
3 years ago
Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of
sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass  2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁  = I₂ ω₂  ,  ω₁ and   ω₂ are angular velocities initially and finally .

I₁ / I₂     =  ω₂ / ω₁

I₁ / I₂     =  T₁ / T₂  , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3  x 2.3 x 10¹⁹ / M

= 1 + 5 / 3  x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0
3 years ago
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