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Marizza181 [45]
2 years ago
10

Find the frequency, if the amplitude of a 3000g object in simple harmonic motion is 1000cm and the maximum speed of the object i

s 5.00 m/s
3.14 Hz
7.96 102 Hz
0 278 × 10 2 HZ
O 0.500 Hz
2.00 Hz
Physics
1 answer:
Oksana_A [137]2 years ago
7 0

Answer:

A = 10 m     amplitude

m = 3 kg     mass of object

Vm = 5 m/s

w A = Vm      where w = omega

w = 2 * pi * f

2 * pi * f  10 = 5

f = 5 / (20 * pi)  = .0796 / sec

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Explanation:

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Newton’s law is the answer
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Is the change in velocity divided by the time needed for the change to occur.
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Answer: yes

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

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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

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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

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(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)

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(d)

A turtle

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Size =  22. cm = 0.22 m

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λ= h/(mv)

λ= 6.62607*10^-9/ (  0.71* 0.028)

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