All categories

2 years ago

Which are methods of reducing exposure to ionizing radiation? Check all that apply.

2 answers:

8 0

<h2>Answer:Answer: using a lead apron during medical X-rays, using a short burst of X-rays to take an image, taking an image of the smallest possible area</h2>

Ionizing radiation is a type of energy released by atoms in the form of electromagnetic radiation (gamma rays or X-rays) or particles (alpha and beta particles or neutrons).

This is a product of the spontaneous disintegration of atoms, where the surplus energy emitted is a form of ionizing radiation.

In this context, people are exposed daily to ionizing radiation, both of natural origin (for example radon emanating from rocks and soil or cosmic rays that go through the Earth) or human, as for example by the generation of nuclear energy or for the medical use of radiation for diagnostic or therapeutic purposes.

In the latter case, the most common artificial sources of ionizing radiation are medical devices, such as X-ray machines.

In this regard, it should be noted that X-rays are dangerous, because they contain enough energy to trigger mutations in organs and tissues and cause cancer (the damage caused depends on the dose received or absorbed). That is why people who undergo X-rays should be exposed to a brief burst of this radiation to capture the image on a photographic film, as well as avoid taking very large areas of the body and the staff that works with them must be protected wearing a lead apron and be placed behind a thick concrete wall as these materials prevent these rays from passing.

Dominik [7]2 years ago

5 0

Answer: (1,4,5)

Explanation:

1, 4, 5, are the answers on edg

You might be interested in

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

$V_{b} - V_{a}$ = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

V_{b} - V_{a} = - ∫ E ds

We substitute

V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

V_{b} - V_{a} = - V_{b} - V_{a}

V_{a} - V_{b} = 89.3

3 0

2 years ago

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t

Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

$\int\limits {F} \, dx$

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

$\frac{dV}{dt} = a = 2.6$

So we have:

$v = 2.6t$

Then:

$\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt$

3) Finally, we replace everything:

$\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt$

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

$P = \frac{W}{t}$

P = 161.9638/4.7 = 34.46 W

8 0

2 years ago

Which of the following relationships must be true according to the laws of series and parallel connections? (Select only relatio

tester [92]

Answer:

Explanation:

a False

b True

c True

d False

e False

f False

g True

h False

i False

4 0

2 years ago

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

Solve the equation x=3logy2 for y.

melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0

2 years ago

Read 2 more answers