Answer:
the answer is A
Explanation:
I dont really have an explanation but hope it helps
Answer : (C) "Higher frequencies have larger spaces between lines".
Explanation:
In Young's experiment, the condition for constructive interference is given by :
.........(1)
n is order or number of lines observed
d is distance between slits
is the angle between the path and the line from screen to the slits.
We also know that, 
or

where,
c is the speed of light
is frequency
is wavelength
So, equation (1) turns into


So,

or
Higher frequencies have larger spaces between line.
So, correct option is (C).
The half-life of the reaction is 50 minutes
Data;
- Time = 43 minutes
- Type of reaction = first order
- Amount of Completion = 45%
<h3>Reaction Constant</h3>
Let the initial concentration of the reaction be X
The reactant left = (1 - 0.45) X
= 0.55 X
= X
For a first order reaction

<h3>Half Life </h3>
The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

Substitute the values

The half-life of the reaction is 50 minutes
Learn more on half-life of a first order reaction here;
brainly.com/question/14936355
Answer:
The carrying capacity of this population would be 125 we know this because we see that this number occur multiple times and seems to be the tipping point after which the number of the population always go down