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Kamila [148]
3 years ago
13

Finley is trying to transmit a sound wave that will travel continuously, but he is frustrated by the excess noise. What kind of

signal is Finley sending?
Physics
2 answers:
Georgia [21]3 years ago
5 0

Answer:

Analog Signal

Explanation:

An analog signal is a continuous signal in which both quantities are analogous to time, that is they change with time.

The noise Finley is observing if the electronic noise and distortion which is introduced by communication channel ( media in which sound is travelling).

During the transmission of analog signal, noise is the biggest disadvantage as during this process the unwanted noise results into generation loss ( a irreversible loss ) resulting a degrade into Signal to Noise ratio.

NikAS [45]3 years ago
3 0
The answer is Analog
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A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

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