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3 years ago

5

Radioisotopes often emit alpha particles, beta particles, or gamma rays. The distance they travel through matter increases in or

der from alpha to gamma. Each radioisotope has a characteristic half-life, which is the time needed for half of a sample of radioisotope to undergo nuclear decay. Which quality is desirable for a radioisotope that is used for medical imaging of a specific organ

1 answer:

Natalija [7]3 years ago

5 0

Answer:

the distance/penetrative power

Explanation:

this is important since there are different electromagnetic waves used to observe different parts like some are used to detect tumours and growths and others fractures so to detect eg fractures the radiation must be highly penetrative to penetrate skin and flesh to reach the bones but it's half-life will

only matter when you look at environmental issues and stuff

You might be interested in

A bowling ball encounters a 0.760-m vertical rise on

matrenka [14]

Answer:1.26 m/s

Explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com $I=\frac{2}{5}mr^2$

Initial Energy

$E_i=\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2(\omega =\frac{u}{r})$

Final Energy

$E_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh$

Equating energy as no energy loss take place

$E_i=E_f$

$\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh$

$\frac{1}{2}mu^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{u}{r}\right )^2=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{v}{r}\right )^2+mgh$

m term get cancel

$\left ( \frac{u^2}{2}\right )+\left ( \frac{2u^2}{10}\right )=\left ( \frac{v^2}{2}\right )+\left ( \frac{2v^2}{10}\right )+gh$

$\frac{7}{10}u^2=\frac{7}{10}v^2+gh$

$v^2=3.5^2-\frac{10}{7}\times 9.81\times 0.76$

$v=\sqrt{1.6}=1.26 m/s$

7 0

2 years ago

A 51.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each

chubhunter [2.5K]

Answer:

Pressure = 9.94 x 10⁶ Pascals

Explanation:

given data

mass = 51 kg

radius = 0.400 cm

solution

we know Pressure that is express as here

Pressure = total force on an area ÷ the area of the area .................1

and

Force is the woman's weight so weight will be

Weight = mass × gravity .................2

put here value

Weight = 51 × 9.8 m/s²

Weight = 499.8 Newtons

and

Area of a circle of bottom of the heel = (π) × (radius)² ...................3

put here value

Area = (π) × (0.40 cm)²

Area = 0.502654 cm²

Area = 0.0000502654 m²

and

now we put value in equation 1 we get

Pressure = force ÷ area

Pressure = 499.8 ÷ 0.0000502654

Pressure = 9943221.381 N/m²

Pressure = 9.94 x 10⁶ Pascals

6 0

2 years ago

A rod of length L and electrical resistance R moves through a constant uniform magnetic field B with rightwards harpoon with bar

NNADVOKAT [17]

Answer:

0 N

Explanation:

Magnetic force for charges moving in wire is given by

F = qvB sin θ

It can also be rewritten for a current carrying wire as

F = ILB sin θ

IL = product of the length of wire and magnitude of the current vector

B = magnetic field.

θ = the angle between the direction of current flow and the magnetic field. For this question, the direction for the two are both parallel; θ = 0°

F = IL B sin 0 = 0 N

Hope this Helps!!!

4 0

2 years ago

a boat sails along the shore. to an observer, the boat appears to move at a speed of 3m/s, and a man on the boat walking foward

lina2011 [118]

It is 6.7s. you need to slove the promble

6 0

2 years ago

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0×1011 solar masses. A star orbiting on

butalik [34]

Answer:

a) $T = 26.147\times 10^{7}\,y$ , b) $1.520\times 10^{13}$ solar masses

Explanation:

Let suppose that galaxy can be treated as a puntual mass.

a) The acceleration experimented by the star is:

$a_{r} = G\cdot \frac{M}{r^{2}}$

$a_{r} = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (8.0\times 10^{11})\cdot\frac{\left(1.989\times 10^{30}\,kg\right)}{\left[(6.0\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)\right]^{2}}$

$a_{r} = 3.296\times 10^{-10}\,\frac{m}{s^{2}}$

The angular speed of the star is:

$\omega = \sqrt{\frac{a_{r}}{r} }$

$\omega = \sqrt{\frac{3.296\times 10^{-10}\,\frac{m}{s^{2}} }{(6.0\times 10^{4})\cdot (9.461\times 10^{15}\,\frac{m}{ly} )} }$

$\omega = 7.62\times 10^{-16}\,\frac{rad}{s}$

The period is:

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{7.62\times 10^{-16}\,\frac{rad}{s} }$

$T = 8.246\times 10^{15}\,s$

$T = 26.147\times 10^{7}\,y$

b) The period is:

$T = 6\times 10^{7}\,y$

$T = 1.892\times 10^{15}\,s$

The angular speed is:

$\omega = \frac{2\pi}{1.892\times 10^{15}\,s}$

$\omega = 3.321\times 10^{-15}\,\frac{rad}{s}$

The acceleration experimented by the star is:

$a_{r} = \left(3.321\times 10^{-15}\,\frac{rad}{s} \right)^{2}\cdot (6\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)$

$a_{r} = 6.261\times 10^{-9}\,\frac{m}{s^{2}}$

The mass of the galaxy is:

$M = \frac{a_{r}\cdot r^{2}}{G}$

$M = \frac{\left(6.261\times 10^{-9}\,\frac{m}{s^{2}} \right)\cdot \left[(6.0\times 10^{4}\ly)\cdot (9.461\times 10^{15}\,\frac{m}{ly} )\right]^{2}}{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }$

$M = 3.023\times 10^{43}\,kg$

Which is equal to $1.520\times 10^{13}$ solar masses.

5 0

2 years ago