the answer is 5 m up AKA "B"
Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center

Second minima from center

The distance between the places where the intensity is zero due to the double slit effect



Put the value into the formula



Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
go to the link quizzlet it will give you tha answer
Explanation:
Using your periodic table if you look at it 3-11 are tansition metals so the horizontal Group Number will help if the group number has to digits just remove the one so if it were to be 13, the valence would be 3, if it were 14 the valence would be ,4 if it were 15, the valence would be 5, if it were 16 the valence would be 6, if it were 17 the valence would be 7 if it were group 18 the valence would be 8 so if anymore help needed to explain hit me up
Answer:

Explanation:
Let's use the decay equation.

Where:
- A is the activity at t time
- A₀ is the initial activity
- λ is the decay constant
We know that 
So we have:




Therefore, the half-life of the source is 6 hours.
I hope it helps you!