Um tubo de raios-X é um tubo de vácuo que converte a energia elétrica em raios-X. A disponibilidade dessa fonte controlável de raios-X criou o campo da radiografia, a imagem de objetos parcialmente opacos com radiação penetrante. Em contraste com outras fontes de radiação ionizante, os raios X são produzidos apenas enquanto o tubo de raios X estiver energizado. Os tubos de raios-X também são utilizados em scanners de tomografia computadorizada, scanners de bagagem de aeroportos, cristalografia de raios-X, análise de materiais e estrutura e para inspeção industrial.
To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation

Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by

Then replacing our values we have that


We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,



The uncollided flux density at the outer surface of the tank nearest the source is 
The resultant force on the object is
∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N
which has a magnitude of
<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N
By Newton's second law, the acceleration has magnitude <em>a</em> such that
<em>F</em> = <em>m a</em>
10 N = (2 kg) <em>a</em>
<em>a</em> = (10 N) / (2 kg)
<em>a</em> = 5 m/s²
so the answer is B.
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 (
)
β₂ - β₁ = 10
log \frac{I_2}{I_1} =
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m
Answer:
A (2066,6 N)
Explanation:
Use the Work formula
62.000J = F . 30
62.000/30 = 2066,6 N
The amout of time it took to move the rock doesn´t matter at all.
It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.