The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter 
:
The total momentum is the sum of the momentums. The initial situation is the following:
(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).
So, at the beginning, the total momentum is
At the end, we have
(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)
After the impact, the total momentum is
Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for 
, and you'll have the final velocity of the 5-kg object.
<u>Mechanics</u> is the branch of physics which deals with the study of motion of material objects.
<u><em>Divisions</em></u>
There are three major division of mechanics
Statics
Kinematics
Dynamics.
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 / 
- 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.
Answer:
Thorium 227 (also known as Radioactinium)
