Answer:
k = 3.5 N/m
Explanation:
It is given that the time period the bob in pendulum is the same as its time period in spring mass system:
where,
k = spring constant = ?
g = acceleration due to gravity = 9.81 m/s²
m = mass of bob = 125 g = 0.125 kg
l = length of pendulum = 35 cm = 0.35 m
Therefore,
<u>k = 3.5 N/m</u>
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.
<h3>How to calculate the photon energy?</h3>
In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.
Mathematically, the Planck-Einstein equation can be calculated by using this formula:
E = hf
<u>Where:</u>
In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.
Read more on photons here: brainly.com/question/9655595
#SPJ1
Hello,
The answer is "universe, Milky Way, clusters, stars, planets".
Reason:
The universe would be the biggest because it has all the galaxy's, starts, clusters, and planets into one. Then it would be Milky Way because this is a galaxy that contains: stars, planets, and clusters. Then it would be clusters because that contains stars, or planets in one group. Then be stars because stars are bigger than planets. Then it would be planets. Therefore the order should go like this: <span>Milky Way, universe, planets, clusters, and stars.
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit</span>
