Answer:
What is the power of focus from the eye when a subject looks from 20 to 500 from its eye?
Explanation:
Is that your question?
Answer: An equation is missing in your question below is the missing equation
a) ≈ 8396
b) 150 nm/k
Explanation:
<u>A) Determine the number of Oscillators in the black body</u>
number of oscillators = 8395
attached below is the detailed solution
<u>b) determine the peak wavelength of the black body </u>
Black body temperature = 20,000 K
applying Wien's law / formula
λmax = b / T ------ ( 1 )
T = 20,000 K
b = 3 * 10^6 nm
∴ λmax = 150 nm/k
The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.
The fragment of an asteroid or any interplanetary material is known as A. METEROID
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is: