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4 years ago

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Part A A conducting sphere is charged up such that the potential on its surface is 100 V (relative to infinity). If the sphere's

radius were twice as large, but the charge on the sphere were the same, what would be the potential on the surface relative to infinity?

1 answer:

Mars2501 [29]4 years ago

8 0

Answer:

$V_{2}=\frac{V_{1}}{2}$

Explanation:

The potential of a conducting sphere is the same as a punctual charge,

First sphere:

$V_{1}=k*q_{1}/r_{1}$ (1)

Second sphere:

$V_{2}=k*q_{2}/r_{2}$ (2)

But, second sphere's radius is twice first sphere radius, and their charges the same:

$q_{1}=q_{2}$

$r_{2}=2*r_{1}$

If we divide the equations (1) and (2), to solve V2:

$V_{2}=V_{1}*r_{1}/r_{2}=V_{1}/2$

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1 year ago

Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba

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- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is $\frac{1}{2} mv^2$ . Therefore, the total kinetic energy before the collision is
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the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
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But the final kinetic energy, Kf, is also
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$mv_2^2 = mv^2$

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3 years ago

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Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is <u>3.92 m/s</u>.

Explanation:

Let's use <u>Newton's 2nd Law</u> to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

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The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
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The Ferris wheel's tangential velocity is 3.92 m/s.

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