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Likurg_2 [28]
4 years ago
9

Tungsten, W-181, is a radioactive isotope with a half life of 121 days. If a medical lab purchases 24 kg of W-181, how much will

be left after 1 year?
Physics
2 answers:
luda_lava [24]4 years ago
8 0

1 year = (365 / 121) = 3.02 half-lifes.  Let's call it 3 .

The amount of radioactive isotope remaining after 3 half-lifes is

(1/2) x (1/2) x (1/2) = 1/8 

A year after the medical lab received the 24 kg of W-181, 
there will still be 24 kg of stuff in the container. 
But only 3 kg of it will still be W-181.  The other 21 kg will be
whatever substances W-181 becomes when it decays.

Sadly, even the 3 kg of good stuff won't be usable anymore ...
it'll be thoroughly mixed with the 21 kg of junk.  It would be harder
and more expensive to try and separate them than to buy a new
can of pure W-181, and USE it before 7/8 of it has deteriorated.
WITCHER [35]4 years ago
7 0

3kg if you using usatestprep

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When voltage sources are connected in series, the total voltage is equal to the algebraic sum of the individual voltages?
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A small ball (mass of 0.5 kg) and a large brick (mass of 10 kg) are dropped from a height of 5 meters. Both items are released a
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3 years ago
Read 2 more answers
Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.
romanna [79]

Answer:

(a): \rm -5.627\times 10^3\ N.

(b):  \rm 7.626\times 10^2\ N.

Explanation:

<u>Given:</u>

  • Charge on one sphere, \rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.
  • Charge on second sphere, \rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.
  • Separation between the spheres, \rm r=3.59\ cm = 3.59\times 10^{-2}\ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}

where,

k is called the Coulomb's constant, whose value is \rm 9\times 10^9\ Nm^2/C^2.

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, \rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.

The new charges on both the spheres are equal and given by

\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.

The magnitude of the force that each sphere now experiences is given by

\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.

7 0
3 years ago
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
vredina [299]

Answer:

a) v_1=5.5172\ m.s^{-1}

b) v_2=-1.5152\ m.s^{-1}

c) v_3=4.6154\ m.s^{-1}

d) v_{avg}=3.8462\ m.s^{-1}

Explanation:

Given:

  • distance down the field in the first interval, d_1=16\ m
  • time duration of the first interval, t_1=2.9\ s
  • distance down the field in the second interval, d_2=-2.5\ m
  • time duration of the second interval, t_2=1.65\ s
  • distance down the field in the third interval, d_3=24\ m
  • time duration of the third interval, t_3=5.2\ s

a)

velocity in the first interval:

v_1=\frac{d_1}{t_1}

v_1=\frac{16}{2.9}

v_1=5.5172\ m.s^{-1}

b)

velocity in the second interval:

v_2=\frac{d_2}{t_2}

v_2=\frac{-2.5}{1.65}

v_2=-1.5152\ m.s^{-1}

c)

velocity in the third interval:

v_3=\frac{d_3}{t_3}

v_3=\frac{24}{5.2}

v_3=4.6154\ m.s^{-1}

d)

We know that the average velocity is given as the total displacement per unit time.

v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}

v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}

v_{avg}=3.8462\ m.s^{-1}

3 0
3 years ago
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