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BlackZzzverrR [31]
3 years ago
6

Two massive particles with identical charge are launched into the uniform field between two plates from the same launch point wi

th the same velocity. They both impact the positively charged plate, but the second one does so four times as far as the first. What sign is the charge? What physical difference would give them different impact points (quantify as a percent)? How does this compare to the gravitational projectile motion case?
Physics
1 answer:
Svetach [21]3 years ago
6 0
<h2><em><u>A</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em>:</em></h2>

<h2>The interaction between charged object is a non-contact force that acts over some distance of separation.Charge,charge and distance.Every <em>e</em><em>l</em><em>e</em><em>c</em><em>t</em><em>r</em><em>i</em><em>c</em><em>a</em><em>l</em><em> </em><em>i</em><em>n</em><em>t</em><em>e</em><em>r</em><em>a</em><em>c</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>i</em><em>n</em><em>v</em><em>o</em><em>l</em><em>v</em><em>e</em><em>s</em><em> </em><em>a</em><em> </em><em>f</em><em>o</em><em>r</em><em>c</em><em>e</em><em> </em><em>t</em><em>h</em><em>a</em><em>t</em><em> </em><em>h</em><em>i</em><em>g</em><em>h</em><em>l</em><em>i</em><em>g</em><em>h</em><em>t</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>m</em><em>p</em><em>o</em><em>r</em><em>t</em><em>a</em><em>n</em><em>c</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>r</em><em>e</em><em>e</em><em> </em><em>v</em><em>a</em><em>r</em><em>i</em><em>a</em><em>b</em><em>l</em><em>e</em><em>s</em></h2>

er

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How much time would it take for an object to fall 4.7 meters
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Answer:

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4 0
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A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3
shusha [124]

Answer:

a

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Explanation:

From the question we are told that

  The initial position of the particle is  x_1 = 0.180 \ m

  The initial  velocity of the particle is  u = 0.060  \  m/s

  The acceleration is   a = -0.380 \  m/s^2

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=>  v = 0.060 + (-0.380 * 3.80)

=>  v = -1.384 \ m/s

Generally from kinematic equation

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Here s is the distance covered by the particle, so

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=>  s = -2.5156 \ m

Generally the final position of the particle is  

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Answer:

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Let's use the initial data to calculate the spring constant

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Reduscate to the English system

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Now we can use Newton's second law

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        a = - 1.2  0.5  / 0.009375

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Please Help
satela [25.4K]
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