Answer:
The frequency of the photon is
.
Explanation:
Given that,
Energy
We need to calculate the energy
Using relation of energy

Where,
= energy spacing


Put the value of h into the formula


Hence, The frequency of the photon is
.
The object that goes through chemical change, changes completely to where you can not change it back to its original form. Physical change you can undo
Answer:
The beam of light is moving at the peed of:
km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity,
(1)
Now, in the right angle in the given fig.:

Now, differentiating both the sides w.r.t t:

Applying chain rule:


Now, using
and y = 1 in the above eqn, we get:

Also, using eqn (1),


Answer: D. ➡️⬅️
Explanation: I just knew the answer ;)
Venus has a dense atmosphere of mostly carbon dioxide. <em>(D)</em>
A, B, and C are false statements.