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max2010maxim [7]
3 years ago
9

A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statements describe the charging method

shown? Check all that apply.
Physics
2 answers:
vova2212 [387]3 years ago
6 0
<span><em>I just took the test and the following are the correct answer:</em>

</span>✔ The spheres develop opposite charges.✔ Electrons move from Sphere A to Sphere B.<span>✔</span> The spheres are charged through induction.

<em>Hope This Helps.</em>
grin007 [14]3 years ago
3 0
<h3>Answer;</h3>
  • <em>The spheres develop opposite charges. </em>
  • <em>Electrons move from Sphere A to Sphere B. </em>
  • <em>The spheres are charged through induction.</em>
<h3><u>Explanation;</u></h3>
  • <u><em>When a negatively charged rod is placed near two neutral metal spheres, the spheres will develop opposite charges, because the neutral metal spheres have both negative and positive charges. </em></u>From the basic law of electrostatics unlike charges attracts and like charges repel.
  • Thus, <em><u>the sphere will develop opposite charges, electrons will move from Sphere A to sphere B,</u></em> hence we say that the spheres will be charged by induction such that sphere A will acquire a positive charge while sphere B will acquire negative charge.
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Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

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depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1  and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

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3 years ago
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Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

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Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

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Replacing by these values, and solving for v, we have:

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b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

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In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

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\Large{\red{\bf{\blue{\dag} Answer:-}}}

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