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3 years ago

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What is the difference between hard x-rays and soft x-rays? I mainly need to know what soft-rays are used for. :) (Can't find th

e right topic for this question)

1 answer:

bixtya [17]3 years ago

7 0

X-rays with high photon energies (above 5–10keV, below 0.2–0.1nm wavelength) are called hard X-rays, while those with lower energy (and longer wavelength) are called soft X-rays. Due to their penetrating ability, hard X-rays are widely used to image the inside of objects, e.g., in medical radiography and airport security. By contrast, soft X-rays are easily absorbed in air; the attenuation length of 600 eV (~2nm) X-rays in water is less than 1 micrometer.

hope this helps

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a boy standing by a lake sees a fish in the pond and tries to thrust a spear into it he will success or not

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Answer:

yes

Explanation:

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Answer:

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Explanation:

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A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform e

nalin [4]

Answer:

a

$\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

b

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

$r = \frac{0.03}{2}$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^{-4 } \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

So

$\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

So

$\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0$

=> $\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }$

=> $\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }$

=> $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

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3 years ago

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Naddik [55]

Answer:

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Explanation:

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3 years ago

An astronaut travels to a star system 6.5 ly away at a speed of 0.80c . Assume that the time needed to accelerate and decelerate

gtnhenbr [62]

Answer:

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Explanation:

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Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:

1.0t=6.5 which is just t=6.5

so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.

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