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2 years ago

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An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

arge density σ = -0.35 μC/m2. A thin wire, with linear charge density λ = 1.1 μC/m, is inserted along the shells' axis. The shell and the wire do not touch and these is no charge exchanged between them.a) What is the new surface charge density, in microcoulombs per square meter, on the inner surface of the cylindrical shell?b) What is the new surface charge density, in microcoulombs per square meter, on the outer surface of the cylindrical shell?c) Enter an expression for the magnitude of the electric field ou

1 answer:

gayaneshka [121]2 years ago

8 0

Answer:

Explanation:

Solution is in the picture attached

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The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?

liberstina [14]

That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum"; sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear". In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate; then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear"; it's a long skinny chamber, called the "cochlea", wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.

-- there are two complete copies of this whole structure ... one on each side of your head.

6 0

2 years ago

Read 2 more answers

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

Focus groups are one of the most widely used ________ methods to gain greater understanding of a current problem or to develop p

Zigmanuir [339]

Answer:

Exploratory

Explanation:

<u>Focus groups</u>

It is a small group of 8-12 respondents guided by a moderator through a thorough debate on a specific subject or idea.It's great for generation of ideas, brainstorming, insight into motives, attitudes, and perceptions. it can show likes, dislikes,emotional requirements and prejudices.

Exploratory methods are used to gain initial insights that could pave the way for further investigation.

Some of the exploratory methods are focus groups, Key informant,case studies,secondary data and observational data.

8 0

2 years ago

A radiator rests snugly on the floor of a room when the temperature is 10 oC. The radiator is connected to the furnace in the ba

s344n2d4d5 [400]

Answer: 10

Explanation subtract

8 0

2 years ago

11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is finge

Anna35 [415]

Answer:

The frequency is $f_n = 257.1 \ Hz$

Explanation:

From the question we are told that

The third harmonic frequency of the tight guitar string is $f_3 = 540 \ Hz$

Let the original length be L

Then the length at which it is fingered is 0.7 L

Generally the fundamental is mathematically represented as

$f = \frac{v_s}{ 2L}$

Now when it finger at 70% it original length is

$f_n = \frac{v}{2 * (0.7 L)}$

$f_n = \frac{v}{1.4 L}$

Here v the velocity of sound

So

$\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }$

Also the fundamental frequency for the original length can also be represented as

$f = \frac{f_3}{3}$

substituting values

$f = \frac{540}{3}$

$f = 180 \ Hz$

So

$\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }$

=> $f_n =\frac{180}{0.7}$

=> $f_n = 257.1 \ Hz$

3 0

2 years ago