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Likurg_2 [28]
3 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]3 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]3 years ago
7 0

3kg if you using usatestprep

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Item 8
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Answer:isotopes

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8 0
3 years ago
1. What could be the energy of a flying aeroplane?<br><br> 2. Is work the same as power?
Doss [256]

Explanation:

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E = ½ m v2 + mgh

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7 0
3 years ago
A cylindrical tank of methanol has a mass of 70 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific
dem82 [27]

Answer:

Weight=686.7N, \rho=933kg/m^{3}, S.G.=0.933, F=17.5N

Explanation:

So, the first value the problem is asking us for is the weight of methanol. (This is supposing there is a mass of methanol of 70kg inside the tank). We can find this by using the formula:

W=mg

so we can substitute the data the problem provided us with to get:

W=70kg(9.81m/s^{2})

which yields:

W=686.7N

Next, we need to find the density of methanol, which can be found by using the following formula:

\rho=\frac{m}{V}

we know the volume of methanol is 75L, so we can convert that to m^{3} like this:

75L*\frac{0.001m^{3}}{1L}=0.075m^{3}

so we can now use the density formula to find our the methanol's density, so we get:

\rho=\frac{m}{V}

\rho=\frac{70kg}{0.075m^{3}}

\rho=933.33kg/m^{3}

Next, we can us these values to find the specific gravity of methanol by using the formula:

S.G.=\frac{\rho_{sample}}{\rho_{H_{2}O}}

when substituting the known values we get:

S.G.=\frac{933.33kg/m^{3}}{1000kg/m^{3}}

so:

S.G.=0.933

We can now find the force it takes to accelerate this tank linearly at 0.25m/s^{2}

F=ma

F=(70kg)(0.25m/s^{2})

F=17.5N

6 0
3 years ago
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