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ahrayia [7]
1 year ago
6

If the nuclide has a half-life of eight days, what mass of the nuclide remains in the patient at 6:00 p.m. the next day? (assume

no excretion of the nuclide from the body.)
Physics
1 answer:
svlad2 [7]1 year ago
3 0

half life = 8.0 days = 8 * 24 = 192 hr

we know,

half life = 0.693 / K

192 hr = 0.693 / K

K = 3.609 * 10^-3 hr-1

time from 8.00 AM to next day 10.00 pM = 38 hr

Now,

ln (I0 / I) = K * t or

ln (1.6 / I) = 3.609 * 10^-3 * 38 or

I = 1.39 ug

mass of the nuclide remains in the patient = 1.39 ug

We know that half life = 0.693 / K  

                       Half life = 8.0 days (8 ×24)

                   192 hours = 0.693 / K

Consequently, 32 hours pass between 8 AM to 6 PM the following day.

<h3 />

Which nuclide has the longest and shortest half life time?

  • Bismuth-209 (209Bi), a radioisotope of bismuth, has the longest half-life of any radioisotope that undergoes -decay that is currently known (alpha decay). Currently, gold is the heaviest stable monoisotopic element, while lead-208 is the heaviest stable nucleus.
  • Fluorine (9F) has 18 known isotopes, with ages ranging from 13 to 18.
  • Fluorine is a monoisotopic and mononuclidic element since only fluorine-19 is stable and is present in nature in amounts greater than trace amounts.
  • Due to 209Bi's extraordinarily long half-life, it is still possible to treat it as if it were non-radioactive for practically all purposes. It poses a negligible radiation risk because its radioactivity is far lower than that of human flesh.
  • The longest half-life of any radionuclide to have been experimentally identified is that of tellurium-128 (128Te), whose half-life is projected to be 7.7 1024 years through double decay. However, 209Bi still holds the record for alpha decay (double beta decay).

To learn more about nuclide and half life time ,visit:

brainly.com/question/2036074

#SPJ4

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Snezhnost [94]

Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

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i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

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The acceleration due to gravity on the surface of Venus is 8.83 m/s2. An object with a mass of 5.23 kg has what weight on Venus?
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A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is
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Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

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Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

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and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

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3 years ago
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Mandarinka [93]
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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
2 years ago
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