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leva [86]
4 years ago
11

An infinite line of charge with linear density λ1 = -6.6μC/m is positioned along the axis of a thick conducting shell of inner r

adius a = 2.7 cm and outer radius b = 4.5 cm and infinite length. The conducting shell is uniformly charged with a linear charge density λ 2 = 3.1 μC/m.
Physics
1 answer:
tester [92]4 years ago
7 0

Answer:

1210 N/m between the inner and outer edges (repulsive)

252.59 N/m between inner and line charges (attractive)

1210+252.59=1462.59 N/m effective force/metre on the inner edge

Explanation:

f=k*Qq/r^2 was applied, the charges are concentrated at the edges of a ring both inner and the outer were shape edges with uniform charge densities 6.6 micro C/m of distance 4.5-2.7=1.8 cm

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Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
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Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

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On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

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