Frequency division multiplexing operates by dividing the signal into different frequencies
<u>Explanation:</u>
The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.
Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.
Answer:
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Answer:
Relation between initial speed of bullet and height h is given as

Explanation:
As we know that system of block and bullet swings up to height h after collision
So we have

so we have

so speed of the block + bullet just after the impact is given by above equation
Now we also know that there is no force on the system of bullet + block in the direction of motion
So we can use momentum conservation

now we have

“Don't hand that holier than thou line to me” is what the asymptote
said to the removable discontinuity.
The distance between the
curve and the line where it approaches zero as they tend to infinity is the line in the asymptote
of a curve. This is unusual for modern authors but in some
sources the requirement that the curve may not cross the line infinitely often
is included.
The point that does not fit the rest of the graph or is
undefined is called a removable discontinuity. By filling in a single
point, the removable discontinuity can be made connected.
Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg