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2 years ago

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A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener

gy Kair did the air molecules gain from the falling coffee filter? Start from the Energy Principle, and choose as the system the coffee filter, the Earth, and the air.

1 answer:

Mademuasel [1]2 years ago

8 0

The kinetic energy gained by the air molecules is 0.0437 J

<h3 />

Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

$P =mgh$

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

$E= \frac{1}{2} mv^{2}$

$E= \frac{1}{2}$ × $1.4$ × 10⁻³ × (0.7)²

E = 0.000343 J

The kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 - 0.0441

= 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

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Part A:

The proton has a smaller wavelength than the electron.

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Part B:

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Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

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$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

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Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

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Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

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Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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