Answer:
7 m .
Explanation:
For destructive interference
Path difference = odd multiple of λ /2
Wave length of sound from each of A and B.
= speed / frequency
λ = 334 / 172 = 2 m
λ/2 = 1 m
If I am 1 m away from B , the path difference will be
8 - 1 = 7 m which is odd multiple of 1 or λ /2
So path difference becomes odd multiple of λ /2.
This is the condition of destructive interference.
So one meter is the closest distance which I can remain at so that i can hear destructive interference.
Protons, neutrons, and electrons<span> are the three main subatomic particles found in an atom.</span>
Answer:
A) R = (200 i ^ + 100 j ^ + 30k ^) m
, B) L = 223.61 m
, C) R = 225.61 m
Explanation:
Part A
This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.
Let's write the displacements
Descending from the apartment
10 flights of 3 m each, the total descent is 30 m
Z = 30 k ^ m
Offset at street level
L1 = 0.2 i ^ km
L2 = 0.1 j ^ km
Let's reduce everything to the SI system
L1 = 0.2 * 1000 = 200 i ^ m
L2 = 100 j ^ m
The distance traveled is
R = (200 i ^ + 100 j ^ + 30k ^) m
Part B
The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane
L² = x² + y²
L = √ (200² + 100²)
L = 223.61 m
Part C
The magnitude of travel, let's use the Pythagorean theorem for the sum
R² = x² + y² + z²
R = √ (30² + 200² + 100²)
R = 225.61 m