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AlexFokin [52]
4 years ago
13

What is one common source of background radiation?

Physics
2 answers:
boyakko [2]4 years ago
7 0
One common source of background radiation would be radon. It is a radioactive gas<span> that emanates from the ground. </span><span>It is usually found in igneous rock and soil. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
lapo4ka [179]4 years ago
5 0

Answer:

<h2>Cosmic radiation.</h2>

Explanation:

Background radiation is a measure of the ionizing radiation present in the environment. The most common source of this is cosmic radiation, which is formed by high energy sub-particles, which almost at light speed. All these particles come from the Sun and other stars, even from other galaxies.

In addition, another common source is the environmental radioactivity, which is produced by radioactive materials, some naturals, other by humans.

Therefore, the answer that best fit is <em>cosmic radiation. </em>

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A 81 kg block is released at a 3.8 m height. the track is frictionless. the block travels down the track, hits a massless spring
Alenkinab [10]
Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track. 
PE = 81 * 9.8 * 3.8 = 3016.44 J
 Work = 1/2 * 1888 * d^2  
PE = Kinetic energy at the base.
 1/2 * 1888 * d^2 = 3016.44
 d = 1.78 approx 1.8
 F = Ke = 1888 * 1.8 = 3398.4N
8 0
3 years ago
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
A machine shop worker reports the mass of an aluminum cube as 176 g. If one side of the cube measures 4 cm, what is the density
Zarrin [17]

-- Since it's a cube, its length, width, and height are all the same 4 cm .

-- Its volume is (length x width x height) = 64 cm³ .

-- Density = (mass) / (volume)

                = (176 g) / (64 cm³) 

                =         2.75 gm/cm³ .

6 0
3 years ago
If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity
Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

8 0
3 years ago
A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly  y  =  0.0204 \ m

Explanation:

From the question we are told that

  The diameter of the glass tube is  d =  1 \ mm =  0.001 \ m

   The density of glycerin is  \rho =  1260 \ kg /m^3

   The surface tension of the glycerin is \sigma   =  6.3 *10^{-2} \ N /m

The capillary rise of the glycerin is mathematically represented as

       y  =  \frac{4 * \sigma  *  cos (\theta )}{ \rho * g *  d}

substituting value  

       y  =  \frac{4 * 6.3 *10^{-2}  *  cos (0 )}{ 1260 * 9.8 *  0.001}

      y  =  0.0204 \ m

Therefore the height  of the glass tube  the glycerin was able to cover is

y  =  0.0204 \ m  

4 0
4 years ago
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