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Svetradugi [14.3K]
3 years ago
10

When a charged particle moves along a helical path in a uniform magnetic field, which component determines the pitch of the path

? the velocity component perpendicular to the magnetic field vector the velocity component parallel to the magnetic field vector the acceleration component perpendicular to the magnetic field vector the acceleration component parallel to the magnetic field vector the acceleration component radially inward the acceleration component radially outward
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

the velocity component parallel to the magnetic field vector

Explanation:

When a charged particle moves in a helical path, we can decompose its velocity into two parts v_parallel and v_perpendicular to the magnetic field.

Let's analyze which component receives a force

            F = q vxB

the bold letters indicate vectors, in the vector product if the two vectors are parallel the angle is zero and the sin 0 = 0 for which there is no force. therefore the velocity parallel to the field remains constant

If the two vectors are perpendicular, the angle is 90º and the sin 90 = 1, for which there is a force, which has a radial direction and consequently a centripetal acceleration that gives a circular path that does not remove the particle from the magnetic field

When checking the different answers, the correct one is: the velocity component parallel to the magnetic field vector

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A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

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Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

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<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

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\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

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<h2>Question B:</h2><h3>First Ball</h3>

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v^2 = 600.25

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<h2>Question C:</h2><h3>First Ball</h3>

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\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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