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mr_godi [17]
3 years ago
5

D. Ampere

Physics
2 answers:
Naily [24]3 years ago
4 0
Features of the mobilization monkey

KiRa [710]3 years ago
3 0

Answer:

0.5

Explanation:

u must simplyfiy

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(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
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

8 0
3 years ago
Answer <br> A<br> B<br> C<br> D and <br> E
harina [27]

Answer:

what is the question I cannot click the

Explanation:

4 0
3 years ago
A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of ca
larisa86 [58]

Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

8 0
4 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
Q30. When working near some large air cored inductors you misplace a socket and as you cannot find it you
Andre45 [30]

When the piece of metal blundered into the air-core coil, it changed the inductance of the coil.

4 0
3 years ago
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