The answer I'm going with is false
Answer:
They would land at the same time
Explanation:
They would land at the same exact time.
As weird, impossible and unbelievable as it appears. When in a vacuum, every weight, body and material when released from the same height would land on the ground at the same time. This also means that like in the question, a feather and a ball would land at the same time. And just for illustrations as well, a feather and a car would land at the same time as well.
Answer:
The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.
Explanation:

Formula used for the radius of the
orbit will be,
where,
= energy of
orbit
n = number of orbit
Z = atomic number
Here: Z = 1 (hydrogen atom)
Energy of the first orbit in H atom .

Energy of the second orbit in H atom .

Energy of the third orbit in H atom .

Energy of the fifth orbit in H atom .

Energy of the sixth orbit in H atom .

Energy of the seventh orbit in H atom .

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :
1) n = 2 → n = 3
2) n= 5 → n = 6
Energy absorbed when: n = 2 → n = 3


Energy absorbed when: n = 5 → n = 6


1.89 eV > 0.166 eV
E> E'
So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.
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Answer:
The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.
Explanation:
Given:
The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1
The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1
The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1
Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1