<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: 
beryllium-15: 
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:
Where 
is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.
Answer:
Explanation:
<u>Conservation of Momentum
</u>
The total momentum of a system of two particles is
Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now
The momentum is conserved if no external forces are acting on the system, thus
Let's put some numbers in the problem and say
120=120
It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s
Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula
Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then
Answer:
The approximate change in entropy is -14.72 J/K.
Explanation:
Given that,
Temperature = 22°C
Internal energy 
Final temperature = 16°C
We need to calculate the approximate change in entropy
Using formula of the entropy
Where, 
= internal energy
T = average temperature
Put the value in to the formula
Hence, The approximate change in entropy is -14.72 J/K.
