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3 years ago

4 points

1 answer:

4 0

Answer: i may not be 100% accurate but i believe it is iron, there is a fire work that uses iron to shine a certain way and is found on stars.

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When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted

irinina [24]

Answer:

Effective half-time of the tracer is 3.6 days

Explanation:

The formula for calculating the decay due to excretion for the first process is ;

$\frac{dN_1}{dt } = - \lambda _e N_o$

here ;

$N_o$ = initial number of tracers

Then to the second process ; we have :

$\frac{dN_2}{dt } = - \lambda _e N_o$

The total decay is as a result of the overall process occurring ; we have :

$\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}$ ------ (1)

here ;

$\frac{dN_{total}}{dt } = \lambda _{total} N_o$

Putting the values in (1);we have :

$- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})$

$\lambda _{Total} = \lambda_e + \lambda r$

As we also know that:

$\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}$

$\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}$

$t_{1/2}_{effective}} = \frac{54}{15}$

= 3.6 days

5 0

4 years ago

Read 2 more answers

Which of the following best describes applying a force over a period of time?

Mazyrski [523]

Impulse
it’s the only one that makes sense energy is just light and power almost the same thing

7 0

3 years ago

Which of these experiments would make use of qualitative data?

Artyom0805 [142]

The answer is A study of different surfaces to compare ability to repel water. Hope this helps!

8 0

4 years ago

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Wich of yhe following is true A mass drpendent on gravity and weight is dependent on gravity B mass is not dependent on gravity

Diano4ka-milaya [45]

C I believe is the answer

6 0

3 years ago

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What is the strength of an electric field that will balance the weight of a 3.3 g plastic sphere that has been charged to -7.6 n

sp2606 [1]

As the plastic sphere is charged, therefore it experience an electric force when placed in an electric fields and also experiences gravitational force acts downward so the electric force must act upward.

Let $F_{E}$ is electric force and $F_{G}$ is gravitational force.

If these forces are balanced, therefore $F_{E} = F_{G}$

or $\left | q \right |E=mg\\\\\ E=\frac{mg}{\left | q \right |}$

Given, $q=-7.6 nC=-7.6\times10^{-9} C$ and $m=3.3 g= 3.3\times10^{-3} kg$ .

Substituting these values in above equation we get,

$E=\frac{3.3\times10^{-3} kg\times9.8 m/s^{2} }{7.6\times10^{-9} C}\\\\E=4.2\times10^{6} N/C$

Thus, the magnitude of electric field is $4.2\times10^{6} N/C$ .

As the charge is negative, the electric field at the location of the plastic sphere must be pointing downward.

8 0

3 years ago