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3 years ago

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An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin

der. In this problem we consider an imaginary cylinder with radius r around the axis AB where r

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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Which one is made up of one or more molecules?

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Atoms are made up or one or more molecules

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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

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Does mars has a bulge near its equator ?

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Yessir it sure does

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3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

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