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shutvik [7]
3 years ago
7

Assume that the wavelengths of photosyntheticallyactive radiations (PAR) are uniformly distributed at integernanometers in the r

ed spectrum from 675 to 700 nm.a. What are the mean and variance of the wavelength distributionfor this radiation?b. If the wavelengths are uniformly distributed at integernanometers from 75 to 100 nanometers, how do the meanand variance of the wavelength distribution compare to theprevious part? Explain.

Physics
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

Attached is the complete solution:

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A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in
MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

1225 = U² + 400

U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

5 0
3 years ago
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the ener
Stella [2.4K]

Answer:

Reflective

Explanation:

The radiation pressure of the wave that totally absorbed is given by;

P_{abs}= \frac{I}{C}

and While the radiation pressure of the wave totally reflected is given by;

P_{ref}= \frac{2I}{C}

Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.

6 0
3 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
Jacob's family celebrates with him as he jumps with both feet together. Jacob is likely (2 points)
professor190 [17]
Jacob is three years old
3 0
2 years ago
Read 2 more answers
Which characteristic should a good scientific question have?
Brilliant_brown [7]
The best answer to this question is D. a god scientific question should be based on observations instead of just imagination.

Hope this helps.
7 0
2 years ago
Read 2 more answers
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