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1 year ago

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A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is

?

1 answer:

tangare [24]1 year ago

7 0

Answer:

h over 2 dg

Explanation:

brainliest!!!!!!!

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A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$

Thus resistance will be R = ρ $\frac{L}{A}$

where A = π $r^2$

so,

$R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}$

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2 years ago

Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani

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d not joining FRA and joining HOSA INSTEAD

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2 years ago

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The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

svlad2 [7]

Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

Explanation: On Edgenuity!!

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2 years ago

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

Elis [28]

Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$

$R = 1.71 * 10^{-4}$ Ω

$R = 171$ μΩ

FOR (b)

for mid point x = L/2

E = p(x)L

for x = L/2

$p(L/2) = a+b(L/2)^2$

for given current I = 1.75 A

so electric field

$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

we get;

$E = 1.7 *10^{-4} V/m$

(here * stand for multiplication )

c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

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2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the

sergeinik [125]

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3 years ago

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