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il63 [147K]
3 years ago
10

Assume that the wavelengths of photosynthetically active radiations (PAR) are uniformly distributed at integer nanometers in the

red spectrum rom 675 to 700 nm. 1. What is the mean of the wavelength distribution for this radiation? 2. What is the variance of the wavelength distribution for this radiation? 3. If instead, the wavelengths are uniformly distributed at integer nanometers from 75 to 100 nanometers, how do the mean and variance of the wavelength distribution compare to the original distribution? Briefly explain.
Physics
1 answer:
Finger [1]3 years ago
7 0

Answer: See explanation

Explanation:

1. What is the mean of the wavelength distribution for this radiation?

Mean = (a+b)/2

= (675 + 700)/2

= 1375/2

= 687.5

2. What is the variance of the wavelength distribution for this radiation?

Variance= {(b-a+1)^2 - 1}/12

= {(700 - 675 + 1)-1}^2/12

= (26^2)-1/12

= 676-1/12

= 675/12

= 56.25

3. If instead, the wavelengths are uniformly distributed at integer nanometers from 75 to 100 nanometers, how do the mean and variance of the wavelength distribution compare to the original distribution?

New Mean = (a+b)/2

= (75 + 200)/2

= 175/2

= 87.5

New Variance= {(b-a+1)^2 - 1}/12

= {(100 - 75 + 1)-1}^2/12

= (26^2)-1/12

= 676-1/12

= 675/12

= 56.25

From the solutions, while the mean differs, the variance (56.25) is thesame due to the fact that variance depends on interval length and the first and second question has same interval length.

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The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

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For 0.25 m the average speed will be:

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For the 0.50 m, the average speed will:

S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg}  = 3.13\rm \  s

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

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Explain the process of making fabric from fibres ​
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3 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

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3 years ago
An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa
Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

f'=f\frac{v\pm v_o}{v\pm v_x}

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

v_x=\frac{f}{f'}(v+v_o)-v

on substituting the values, we get

v_x=\frac{1790}{1780}(343+2.36)-343

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Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0
3 years ago
A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte
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Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}

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So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

6 0
3 years ago
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