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Alla [95]
3 years ago
6

why it is not possible to determine drift velocity of electron by timing its travel through conductor.

Physics
1 answer:
gtnhenbr [62]3 years ago
6 0

The main reason is that electrons are too small to see. 

But even if you could see them, a better reason is the fact that
all electrons are identical.  So if you spotted one electron on its
way into the conductor, and then start watching the other end to
see when your electron gets there, you would see trillions of trillions
of trillions of electrons pouring out over there, and you'd have no way
to tell which electron is the one you're waiting for.

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storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

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