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Anvisha [2.4K]
3 years ago
11

Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of 430nm. Calculate the frequency of this light.
Physics
1 answer:
Ratling [72]3 years ago
7 0

Answer:

6.97 E 16

Explanation:

Frequency is a function of velocity of light to it's wavelength.

Mathematically written as

F = Velocity / wavelength

Velocity of light = 3 x 10^8

Wavelength =430 nm =430 x 10^-9 m

converting wavelength from nanaometer to meter we divide by 10^9

Frequency = (3 x 10^8)/(430 x10^-9) =6.97 E 16

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Ede4ka [16]

Answer:

you didnt put anything

Explanation:

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Which are ways to improve the design of this experiment? Check all that apply.
forsale [732]
-Reduce the sample size so the experiment can be done faster.-Increase the sample size from 6 cups to 12 cups of sand and water.-Use more legible handwriting when recording data.-Use more precise digital thermometers.<span>-Use more precise scales that measure to the hundredth of a gram.</span>
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what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
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Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi
Vedmedyk [2.9K]

Answer:

v = 1.98\ m/s

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m             u₄ = 4.5 m/s

m₅ = m              u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

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5 0
2 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
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