This is true chordates are vertebrae
1 - Skull
2 - Mandible
3 - Scapula
4 - Sternum
5 - Ulna
6 - Radius
7 - Pelvis
8 - Femur
9 - Patella
10 - Tibia
11 - Fibula
12 - Metatarsals
13 - Clavicle
14 - Ribs (rib cage)
15 - Humerus
16 - Spinal column
17 - Carpals
18 - Metacarpals
19 - Phalanges
20 - Tarsals
21 - Phalanges
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
