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3 years ago

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Which of the following is an appropriate receive filter bandwidth for minimizing noise and interference for CW reception?A. 500

Hz
B. 1000 Hz
C. 2400 Hz
D. 5000 Hz

1 answer:

dimulka [17.4K]3 years ago

4 0

Answer: option A= 500 Hz.

Explanation:

The full meaning of CW is continuous waveform or continuous wave. Continuous waveform during wireless telegraph radio transmission is also known as undamped waves. Around 1920, damped wave spark transmitters were replaced by continuous wave vacuum tube transmitters. Continuous waveform is some wave of constant amplitude and frequency.

Continuous waveform consists of a pure sine wave multiplied with a square wave that's either 0 or 1, corresponding to the keying of the carrier. Continuous waveform receive filters with passband around 500 Hz.

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In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,

taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

$f = \frac{1}{2\pi \sqrt{LC} }$

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

$f' = \frac{1}{2\pi \sqrt{2L2C} }$

$f' = \frac{1}{2\pi \sqrt{4LC} }$

$f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}$

$f' = \frac{1}{2} f$

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

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2 years ago

A wind turbine is an example of what kind of device

Liula [17]

Answer:

it’s an example of a generator.

Explanation:

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2 years ago

What would be the indication if there is a high frequency and pitch?

il63 [147K]

Higher pitched sounds produce waves which are closer together than for lower pitched sounds. A smaller triangle or cymbal will make a relatively higher pitch note

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3 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

$2gh = v_f^2 - v_i^2$

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

$(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\$

<u>h = 0.82 m</u>

Now, for the time in air during upward motion we use first equation of motion:

$v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s$

(c)

Now we will consider the downward motion and use the third equation of motion:

$2gh = v_f^2-v_i^2$

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

$2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\$

<u>vf = 7.17 m/s</u>

Now, for the time in air during downward motion we use the first equation of motion:

$v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s$

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

<u>t = 1.14 s</u>

7 0

3 years ago