Number 1 Bc of wood a magnetic
<u>Given:</u>
Initial temperature, T1 = 20 C
Final temperature, T2 = 40 C
<u>To determine:</u>
The temperature change
<u>Explanation:</u>
Convert degree C to Kelvin
Temperature in Kelvin = degree C + 273
T1 = 20 + 273 = 293 K
T2 = 40 + 273 = 313 K
Incremental temperature change = T2 - T1 = 313-293 = 20 K
Ans: The temperature change in kelvin is 20 K
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 = 
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)
0.9715 Fraction of Pu-239 will be remain after 1000 years.
THe Answer is 105 J
hope this helps and please brainliest, I was first :)
