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3 years ago

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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

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

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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A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

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Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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2 years ago

1. Which type of wire would be a better conductor of an electrical current?

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Answer:

1. a

2. b

3. b

Explanation:

1.

Resistance is the property of a conductor to offer resistance to the flow of current. The lower the resistance better is the conductivity of wire.

We know that the resistance of a wire depends on several factor which are inter-connected by an equation as:

$R=\rho.\frac{l}{a}$

where:

R = resistance of the wire

$l =$ length of the wire

$a=$ cross sectional area of the wire

from the above relation we observe that

$R\propto l$

$R\propto \frac{1}{a}$

Also when the temperature of the wire is significantly high then the lattice vibration cause obstruction in the path of the flowing charges and reduce the current flow.

2.

As the collision between the electrons and protons increases the speed of the flow of charges will decrease because the opposite charges attract each other and as we know that electrical current is the rate of flow of charge.

3.

Heating up of wire due to sunlight will cause lattice vibration in the conductor and will obstruct the movement of the charges which build up electric current, hence increasing the resistance of conductivity.

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$V_{e}=\sqrt{\frac{2GM}{R}}$

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$V_{e}$ is the escape velocity

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$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

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$V_{e}=3633.86 m/s \approx 3.63 km/s$

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