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3 years ago

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A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of

the torque on the particle about the origin

1 answer:

NeX [460]3 years ago

8 0

<h2>Question </h2>

A particle, mass 0.25 kg is at a position (<em>-7i + 7j + 5k</em>) m, has a velocity (<em>6i - j + 4k</em>) m/s, and is subject to a force (<em>-5i + 0j - k</em>) N. What is the magnitude of the torque on the particle about the origin?

<h2>Answer:</h2>

47.94Nm

<h2>Explanation:</h2>

The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e

τ = r x F

Given:

r = (-7i + 7j + 5k) m

F = (-5i + 0j - k) N

| i j k |

r x F = | -7 7 5 |

| -5 0 -1 |

r x F = i(-7 - 0) - j(7+25) + k(0+35)

r x F = i(-7) - j(32) + k(35)

r x F = -7i - 32j + 35k

Therefore the torque τ = -7i - 32j + 35k

The magnitude of the torque is therefore;

|τ| = $\sqrt{(-7)^2 + (-32)^2 + (35)^2}$

|τ| = $\sqrt{49 + 1024 + 1225}$

|τ| = $\sqrt{2298}$

|τ| = 47.94Nm

The magnitude of the torque on the particle about the origin is 47.94Nm

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

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Part B:

The proton has a smaller wavelength than the electron.

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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.

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Case for the electron:

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

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

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

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Case for the proton:

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

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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:

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$2KE = mv^{2}$

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

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Then, equation 2 can be used:

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