Answer:
The momentum of the photon is 1.707 x 10⁻²² kg.m/s
Explanation:
Given;
kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J
Kinetic energy is given as;
K.E = ¹/₂mv²
where;
v is speed of the electron
Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s
Answer:
<em>His angular velocity will increase.</em>
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = 'ω'
where
' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia =
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to .
From
'ω' =
ω
since is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.