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Lubov Fominskaja [6]

3 years ago

9

A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further

modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230

1 answer:

Irina-Kira [14]3 years ago

6 0

B I got it right so (;

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How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

3 years ago

what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the un

marta [7]

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin <em>wt</em>

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin <em>wt</em>

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sin<em>wt and w = </em>2 $\pi$ f = $\frac{2\pi }{T}$ rad/sec

3 0

3 years ago

Answers of science book<br><br>

vitfil [10]

Answer: answers of science book is to help you

Explanation: if there are answer in a science book use them there for a reason

6 0

3 years ago

Suppose Q1 (0.0, - 3.0 M, 0.0) = 4.0 nC, Q2 (0.0, 3.0 m, 0.0) = 4.0 nC, and Q3 (4.0 m, 0.0, 0.0) = 1.0 nC.

frozen [14]

Answer:

Explanation:

Force on Q₃ due to charge Q₁

= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ / 5²

= 1.44 x 10⁻⁹ N

Force due to Q₂ will also be 1.44 x 10⁻⁹ N

component of these forces along x axis

-= 2 x 1.44 x 10⁻⁹ cosθ

= 2.88 x 10⁻⁹ x 4/5

= 2.30x10⁻⁹ N along x axis.

The y-component will calcel out.

b ) In this case , Q₁ will repel and Q₂ will attract.

In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹ N . Component along x axis will cancel out.

c ) Electric field in case 1 and case 2 will be

= 2.30x10⁻⁹ / 1 x 10⁻⁹

= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .

5 0

3 years ago

A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 23-lb bar of length L

Sergeeva-Olga [200]

Answer: hello attached below is the missing image the slender weight is different from what is in the question here so I worked with 23-Ib as requested in the question

answer

≈ 12.17 Rad/sec

Explanation:

weight of bullet ( Wb ) = 0.08 Ib

horizontal velocity = 1800 ft/s

Slender(Wr) = 23-Ib bar with

length ( L ) = 30

h = 12 inches

Vro = 0

<u>Calculate the angular velocity of the bar immediately after the bullet becomes embedded </u>

attached below is a detailed solution

6.708 = ( 0.05011 + 0.5011 ) w'

w' = 6.708 / 0.55121 ≈ 12.17 Rad/sec

6 0

3 years ago