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Tasya [4]
3 years ago
14

Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would

require quantum mechanics to describe, that is, would show both particle and wave properties? Which objects would act like everyday objects, and be adequately described by classical mechanics? object quantum or classical? A virus with a mass of 9.4 x 10-17 g, 280. nm wide, moving at 0.50 µm/s. classical quantum A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 37. m/s. classical quantum A mosquito with a mass of 1.0 mg, 6.3 mm long, moving at 1.1 m/s. classical quantum A turtle with a mass of 710. g, 22. cm long, moving at 2.8 cm/s. classical quantum
Physics
1 answer:
HACTEHA [7]3 years ago
8 0

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js  is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle  ------------Quantum

Explanation:

 Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

  = 6.62607*10^-9/4.7*10^-26

  = 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

  =  6.62607*10^-9/4.44*10^-23

  = 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size =  6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  1*10⁻⁶* 1.1)

  =  6.62607*10^-9/1.1*10^-6

  = 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size =  22. cm = 0.22 m

V =  2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  0.71* 0.028)

  = 6.62607*10^-9/0.01988

   = 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

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Answer:

0.75 m

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The bulb and the length of the mirror form a triangle.  The mirror and the illuminated area on the floor form a trapezoid.  If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle.  This triangle and the small triangle are similar.  So we can say:

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Solving for x:

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The large leaves help it survive as they serve as the<u> organ for photosynthesis.</u>

Explanation:

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Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the
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Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, F=1.5\ nN=1.5\times 10^{-9}\ N

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Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

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