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algol [13]
1 year ago
11

A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea

n that a slow neutron appears to be larger to a target particle than does a fast neutron in the sense that the slow neutron has probabilities of being found over a larger volume of space. For a thermal neutron at room temperature of 300 K , find (b) the de Broglie wavelength.
Physics
1 answer:
Vlada [557]1 year ago
6 0

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p =  \sqrt{3RTm}

According to De broglie hypothesis, P = h ÷ λ

Therefore,    λ  = h ÷ \sqrt{3RTm}

                     = 6.6× 10^{-34} ÷  \sqrt{3 \times 1.38 \times  10^{-23}  \times 300 \times 1.6 \times  10^{-27} }

                     = 0.15 × 10^{-9}

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: <u>https://brainly.in/question/6131028</u>

#SPJ4

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The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the
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Answer:

\frac{I}{I_0}=113.68

Explanation:

P = Acoustic power = 63 µW

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Acoustic power

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Ratio

\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

6 0
2 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

2gh = v_f^2 - v_i^2

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

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vi = initial speed = 4 m/s

Therefore,

(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\

<u>h = 0.82 m</u>

Now, for the time in air during upward motion we use first equation of motion:

v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s

(c)

Now we will consider the downward motion and use the third equation of motion:

2gh = v_f^2-v_i^2

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\

<u>vf = 7.17 m/s</u>

Now, for the time in air during downward motion we use the first equation of motion:

v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

<u>t = 1.14 s</u>

7 0
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the distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inch
Paha777 [63]
Direct variation involves ration and proportions, so 

you need to set up the proportion:

<span>11 / 75 = x / 65
 
Cross multiplying:

75x = 11*65

x = (11*65)/75

Solving, we get x = 9.533, </span>

<span>which rounds off to 9.5

Therefore, the spring will stretch up to 9.5 inches with 65 attached.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>
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