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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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Answer:

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Explanation:

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Solution -

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Move the chemistry book across the physics book again. study the thermometer to determine whether there’s a change in the physic
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If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a
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Answer: 1.12 m

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V_{o}=2.2 m/s is the initial velocity

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0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2} (3)

Multiplying all the eqquation by -1 and rearranging:

4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0 (4)

So, since we have a quadratic equation here (in the form of0=at^{2}+bt+c,  we will use the quadratic formula to find  t:  

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}   (5)

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Substituting the known values and choosing the positive result of the equation, we have:  

t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}  

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Now, substituting (6) in (2):

x=(2.2 m/s)(cos 30\°)(0.59 s) (7)

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4 years ago
A square loop of wire has a perimeter of 4.00 mm and is oriented such that two of its parallel sides form a 13.0 ∘∘ angle with t
vivado [14]

Answer:

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θ = angle between the magnetic field and the plane of the area.

a) B = 0.250 T

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Perimeter of a square = 4L

4L = 4.00

L = 1.00 mm = 0.001 m

The area is given as L²

A = (0.001)² = 0.000001 m²

θ = 13°

Φ = BA cos θ

Φ = 0.25 × 0.000001 × cos 13°

Φ = 0.0000002436 Wb = (2.436 × 10⁻⁷) Wb

b) Another loop lies in the same plane but has an irregular shape, resembling a starfish. Its area is three times greater than that of the square loop

Φ = BA cos θ

B = 0.25 T

A = 3 × 0.000001 = 0.000003 m²

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Φ = 0.25 × 0.000003 × cos 13°

Φ = 0.0000007308 Wb = (7.308 × 10⁻⁷) Wb

Hope this Helps!!!

8 0
3 years ago
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