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Xelga [282]
3 years ago
13

A long straight wire has fixed negative charge with a linear charge density of magnitude 4.6 nC/m. The wire is to be enclosed by

a coaxial, thin-walled, nonconducting cylindrical shell of radius 2.2 cm. The shell is to have positive charge on its outside surface with a surface charge density σ that makes the net external electric field is zero. Calculate σ.
Physics
1 answer:
Semmy [17]3 years ago
7 0

Answer: -33.3 * 10^9 C/m^2( nC/m^2)

Explanation: In order to solve this problem we have to use the gaussian law, the we have:

Eoutside =0 so teh Q inside==

the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.

Then we simplify and get

σ= -4.6/(2*π*b)= -33.3 nC/m^2

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The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt
devlian [24]
The original width was 94.71 cm 
<span>The area decreased 33.1% </span>

<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
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<span>X^2 = 10000 * .6 cm </span>
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<span>The length is 2 * 77.46 = 154.92 cm </span>

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<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
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3 years ago
From the list below, which name appears in two different parts of the body? *
Kisachek [45]

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Explanation:

7 0
2 years ago
What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

8 0
3 years ago
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