<h2>
Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life
of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13: ![h_{B-13}=5(10)^{-10}s=0.0000000005s](https://tex.z-dn.net/?f=h_%7BB-13%7D%3D5%2810%29%5E%7B-10%7Ds%3D0.0000000005s)
beryllium-15: ![h_{B-15}=2(10)^{-7}s=0.0000002s](https://tex.z-dn.net/?f=h_%7BB-15%7D%3D2%2810%29%5E%7B-7%7Ds%3D0.0000002s)
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
![h_{B-15}=X.h_{B-13}](https://tex.z-dn.net/?f=h_%7BB-15%7D%3DX.h_%7BB-13%7D)
Where
is the amount we want to find:
![X=\frac{h_{B-15}}{h_{B-13}}](https://tex.z-dn.net/?f=X%3D%5Cfrac%7Bh_%7BB-15%7D%7D%7Bh_%7BB-13%7D%7D)
![X=\frac{2(10)^{-7}s}{5(10)^{-10}s}](https://tex.z-dn.net/?f=X%3D%5Cfrac%7B2%2810%29%5E%7B-7%7Ds%7D%7B5%2810%29%5E%7B-10%7Ds%7D)
Finally:
![X=400](https://tex.z-dn.net/?f=X%3D400)
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.
QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.
ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
The energy of moving electrical charges is Electrical energy
Hope its the answer you are finding and hope it helps....
Answer:
A-500 N
Explanation:
The computation of the tension in the chain is shown below
As we know that
F = ma
where
F denotes force
m denotes mass = 7
And, a denotes acceleration
Now for the acceleration we have to do the following calculations
The speed (v) of the hammer is
v = Angular speed × radius
where,
Angular seed = 2 × π ÷ Time Period
So, v = 2 × π × r ÷ P
v = 2 × 3.14 × 1.8 ÷ 1
= 11.304 m/s
Now
a = v^2 ÷ r
= 70.98912 m/s^2
Now the tension is
T = F = m × a
= 7 × 70.98912
= 496.92384 N
= 500 N
60 days, tell me if I'm correct please.