Giving brainliest to whoever can solve!! Light of wavelength 675 nm passes through a double slit , and makes its first minimum (
m = 1) at an angle of 0.111 deg. What is the separation of the slits , d , in MILLIMETERS ?
1 answer:
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Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - ( )V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.
<h3>What is the Inverse Square Law formula?</h3>The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as
Where
- I₁ = Intensity at distance 1 (W/m²)
- I₂ = Intensity at distance 2 (W/m²)
- d₁ = distance 1 from a light source (m)
- d₂ = distance 2 from a light source (m)
Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?
By using the inverse square law formula, the sun's intensity for an outer planet is
I₂ = 5.55 W/m²
Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².
Learn more about intensity of light here:
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