Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.

Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).

Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.

7 0

3 years ago

You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizont

Rashid [163]

Answer:

47.48 m

Explanation:

The horizontal range of an object can be estimated using the equation below:

$S_{x} =$ $\frac{v^{2}*sin (2\alpha)}{g}$

For the maximum horizontal range to occur, the angle $\alpha$ is 45°. Then, the horizontal range will be:

$S_{x}$ = $\frac{v^{2}*sin 90}{g} = \frac{v^{2}}{g}$

In addition, the velocity (v) can be estimated using the equation below:

v = gt/2

Therefore, the maximum horizontal range is:

$S_{x} = \frac{(gt/2)^{2}}{g}$ = (9.81*4.4/2)^2 / 9.81 = 47.48 m

6 0

3 years ago

Sound travels at a speed of about 340m/s in the air. How long will it take an explosion to be heard from a distance of 20 miles

Goryan [66]

First convert the speed of sound from meters/second to miles/second:

$340\,\dfrac{\mathrm m}{\mathrm s}\cdot\dfrac{3.281\,\mathrm{ft}}{1\,\mathrm m}\cdot\dfrac{1\,\mathrm{mi}}{5280\,\mathrm{ft}}=0.21\,\dfrac{\mathrm{mi}}{\mathrm s}$

Then the time it takes for the sound to travel 20 miles is

$\dfrac{20\,\mathrm{mi}}{0.21\,\frac{\mathrm{mi}}{\mathrm s}}=95\,\mathrm s$

3 0

2 years ago