Book edition 4th edition

Author(s) Ruth W. Chabay, Bruce A. Sherwood

Pages 1135 pages

ISBN 9781118875865

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Short Answer

Question: An electron passes location $(0.02, 0.04, - 0.06) m$ , and $2 μ s$ later is detected at location $(0.02, 1.84, - 0.86) m$ , (1 microsecond is $1 \times 10^{- 6} s$ ). a) What is the average velocity of the electron? b) If the electron continues to travel at this average velocity, where will it be in another $5 μ s$ ?

Answer

a) The average velocity of the ball is $(0, - 900000, 400000) m / s$

b) The electron will be at the position $(0.02, - 0.86, 0.34) m$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of the given data

The given data can be listed below as:

The electron passes the location $(0.02, 0.04, - 0.06) m$ .
The electron has been detected in the location $(0.02, 1.84, - 0.86) m$ .

The electron is detected after $2 μ s$ .

Step 2: Significance of Newton’s first law to calculate the average velocity and position

This law states that an object will continue to move in a uniform motion unless it is resisted by an external object.

Step 3: Determination of the average velocity and distance

a) From Newton’s first law, the formula for the average velocity of the electron can be expressed as:

${\vec{v}}_{a v e r a g e} = (\frac{(x_{B} - x_{A})}{Δ t_{A B}}, \frac{(y_{B} - y_{A})}{Δ t_{A B}}, \frac{(z_{B} - z_{A})}{Δ t_{A B}})$

Here, $t_{A B}$ The time difference of a particle from location A to location B

$x_{A}, y_{A}, z_{A} =$ The positions of the ball at A position in the x, y, and z coordinates

$x_{B}, y_{B,} z_{B} =$ The positions of the ball at B position in the x, y, and z coordinate

Substituting the values localid="1662456388346" $t_{A B} = 2 μ s, x_{A}, y_{A}, z_{A} = (0.02, 0.04, - 0.06) m$ $a n d x_{B}, y_{B,} z_{B} = (0.02, 1.84, - 0.86) m$ in the above equation, we get:

${\vec{v}}_{a v e r a g e} = (\frac{(0.02 - 0.02) m}{2 \times 10^{- 6} s}, \frac{(0.04 - 1.84) m}{2 \times 10^{- 6} s}, \frac{(- 0.06 + 0.86) m}{2 \times 10^{- 6} s}) {\vec{v}}_{a v e r a g e} = (0, - 900000, 400000) m/s$

Thus, the average velocity of the ball is $(0, - 900000, 400000) m/s$ .

b) From Newton’s first law, the equation of displacement of the ball in the x-coordinate can be expressed as:

$x_{C} = x_{B} + v_{a v e r a g e, x} \cdot Δ t_{B C}$

Here, $x_{C} =$ The displacement of the particle in the C point in the x-quadrant,

$t_{B C} =$ The displacement of the particle in the B point in the x-qua

$x_{B} =$ The time difference of a particle from location B to location C

$v_{a v e r a g e, x} =$ The average velocity of the particle in the localid="1662455496234" $x$ quadrant

Substituting the value , localid="1662456406627" $t_{B c} = 3 μ s, v_{a v e r a g e, x} = 0 m/s a n d x_{B} = 0.02 m$ n the above equation, we get:

$x_{C} = 0.02 m + 0 \cdot Δ t_{B C} x_{C} = 0.02 m$

Similarly, the equation of displacement of the ball in the y coordinate can be expressed as:

$y_{C} = y_{B} + v_{a v e r a g e, y} \cdot Δ t_{B C}$

$y_{C} =$ The displacement of the particle in the C point in the y -quadrant,

$y_{B} =$ The displacement of the particle in the B point in the y -quadrant,

$t_{B C} =$ The time difference of a particle from location B to location C

$v_{a v e r a g e, y} \cdot$ The average velocity of the particle in the y -quadrant

Substituting the value $t_{B C} = 3 μ c, v_{a v e r a g e, y} = - 900000 m/s and y_{B} = 1.84 m$ , in the above equation, we get:

$y_{C} = (1.84 m) + (900000 m/s) \times (3μ s) = 1.84 m - 2.7 m = - 0.86 m$

Similarly, the equation of displacement of the ball in the z coordinate can be expressed as:

$z_{C} = z_{B} + v_{a v e r a g e, z} \cdot Δ t_{B C}$

$z_{C} =$ The displacement of the particle in the C point in the z-quadrant,

$z_{B} =$ The displacement of the particle in the B point in the z -quadrant,

$t_{B C} =$ The time difference of a particle from location B to location C

$v_{a v e r a g e, z} =$ The average velocity of the particle in the z quadrant

Substituting the values $t_{B C} = 3 μ s, v_{a v e r a g e, z} = 400000 m/s and z_{B} = - 0.86 m$ in the above equation, we get:

$z_{C} = (- 0.86 m) + (400000 m/s) \times (3s) = - 0.86 m + 1.2 m = 0.34 m$

Thus, the electron will be at the position $(0.02, - 0.86, 0.34) m$ .

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Short Answer

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of Newton’s first law to calculate the average velocity and position

Step 3: Determination of the average velocity and distance

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Matter & Interactions

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Short Answer

Question: An electron passes location (0.02,0.04,-0.06)m, and 2μs later is detected at location (0.02,1.84,-0.86) m, (1 microsecond is1×10-6 s). a) What is the average velocity of the electron? b) If the electron continues to travel at this average velocity, where will it be in another 5μs?

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of Newton’s first law to calculate the average velocity and position

Step 3: Determination of the average velocity and distance

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