The efficiency of a light source is the percentage of its energy input that gets radiated as visible light if some of the blue light in an led is used to cause a fluorescent material to glow the overall efficiency of the LED decreases.
How efficient is LED?
Different wavelengths that correlate to different visible colours are used in LED light therapy. Various shades pierce the skin at different rates.
- Your skin's outermost layer is impacted by blue light.
- Yellow light is more enveloping.
- Red light penetrates your skin more deeply.
- The deepest penetrating light is near-infrared.
Different LED hues have various effects. For instance, according to experts red LED light therapy has the potential to reduce inflammation and boost collagen formation, which declines with age and is crucial for maintaining youthful-looking skin.
Acne-causing bacteria may be destroyed by a blue LED light therapy (P. acnes).
Hence the answer is the overall efficiency of the LED decreases.
Learn more about wavelength here,
brainly.com/question/1263539
# SPJ4
In the offensive role, the players try to get a goal.
In the defensive roll, The players try to protect the goal
Hoped this helped a little :)
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
