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Anestetic [448]
3 years ago
14

Describe three ways you can conserve energy in your own home

Physics
2 answers:
prohojiy [21]3 years ago
7 0
You can turn off the air conditioning
you can keep all electronics off
you can turn off all running water
Mandarinka [93]3 years ago
4 0
Turn lights off, unplug electronics, and use solar energy
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Descreva sucintamente como ocorre a formação do raio x pela ampola ou tubo
Umnica [9.8K]

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.

4 0
3 years ago
A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of
LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

\Psi=4RcE\pi

Where

\Psi =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

I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}

Then replacing our values we have that

I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}

I = 1.3*10^7 Bq

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,

\Psi = \frac{cE}{4\pir^2}

\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}

\Psi = 2.14*10^4 MeV/cm^2.s

The  uncollided flux density at the outer surface of the tank nearest the source is \Psi = 2.14*10^4 MeV/cm^2.s

6 0
2 years ago
4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?
aleksandrvk [35]

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.

4 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

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 \frac{I}{I_o}

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 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 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 = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?
NeX [460]

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.

5 0
3 years ago
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