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Studentka2010 [4]
2 years ago
15

If a receiver is overly selective:

Physics
1 answer:
VMariaS [17]2 years ago
7 0

Answer:

C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.

Explanation:

Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.

If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

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Nookie1986 [14]
The answer is c) 80°c
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2 years ago
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Describe what introspection is.
krek1111 [17]

Answer:

Introspection is a process that involves looking inward to examine one's own thoughts and emotions. ... The experimental use of introspection is similar to what you might do when you analyze your own thoughts and feelings but in a much more structured and rigorous way.

Explanation:

Example of it: The definition of introspection is self-examination, analyzing yourself, looking at your own personality and actions, and considering your own motivations. An example of introspection is when you meditate to try to understand your feelings. noun.

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before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
Virty [35]

The momentum of block B after the collision is -50 kg m/s.

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

p_A + p_B = p'_A + p'_B

where:

p_A = -100 kg m/s is the momentum of block A before the collision

p_B = -150 kg m/s is the momentum of block B before the collision

p'_A = -200 kg m/s is the momentum of block A after the collision

p'_B is the momentum of block B after the collision

Solving for p'_B, we find:

p'_B = p_A + p_B - p'_A = -100 +(-150) -  (-200)=-50 kg m/s

So, the momentum of block B after the collision is -50 kg m/s.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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6 0
2 years ago
A tow truck pulls a 7,800 N car 5m out of a ditch in 6.5 seconds. How much power does the
rosijanka [135]

Answer:

p=6000W

Explanation:

p=w/t

w=f*d=7800*5=39000 J

therefore,

p=39000/6.5=6000 W

6 0
3 years ago
A small metal bead, labeled A has a charge of 25 nC. It is touched to metal bead B, initially neutral, so that the two beads sha
Iteru [2.4K]

Answer:

15 nC and 10 nC

Explanation:

qA + qB = 25 nC = 25 x 10^-9 C

F = 5.4 x 10^-4 N

d = 5 cm = 0.05 m

Use the Coulomb's law

F=\frac{Kq_{A}q_{B}}{d^{2}}

By substituting the values, we get

5.4 \times 10^{-4}=\frac{9 \times 10^{9}q_{A}q_{B}}{0.05^{2}}

qA x qB = 1.5 x 10^-16 C

So, q_{A}\left ( 25 \times 10^{-9}-q_{A} \right )=1.5 \times 10^{-16}

q_{A}^{2}-25 \times 10^{-9}q_{A}+1.5 \times 10^{-16}=0

q_{A}=\frac{25\times 10^{-9} \pm \sqrt{6.25\times 10^{-16}-6 \times10^{-16}}}{2}

q_{A}=\frac {25\times 10^{-9} \pm 5 \times 10^{-9}}}{2}

qA = 15nC or 10nC

So, qB = 10 nC or 15 nC

8 0
3 years ago
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