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marusya05 [52]
2 years ago
9

Source localization in ocean acoustics is posed as a machine learning problem in which data-driven methods learn source ranges d

irectly from observed acoustic data
Physics
1 answer:
Temka [501]2 years ago
6 0

Source localization in ocean acoustics is posed as a machine learning problem in which data-driven methods learn source ranges directly from observed acoustic data: True.

<h3>What is machine learning?</h3>

Machine learning (ML) is also known as artificial intelligence (AI) and it can be defined as a subfield in computer science which typically focuses on the use of computer algorithms, data-driven techniques (methods) and technologies to develop a smart computer-controlled robot that has the ability to automatically perform and manage tasks that are exclusively meant for humans or solved by using human intelligence.

In Machine learning (ML), data-driven techniques (methods) are used to learn source ranges directly from observed acoustic data in a bid to proffer solutions to source localization in ocean acoustics.

In conclusion, a normalized sample covariance matrix (SCM) is constructed and used as the input, especially after pre-processing the pressure that's received by a vertical linear array in Machine learning (ML).

Read more on machine learning here: brainly.com/question/25523571

#SPJ1

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¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Zina [86]

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

distancia=\frac{1 segundo*90 km}{3600 segundos}

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

distancia= 25 metros

<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

6 0
4 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
NASA launches a probe with a mass of 15,000 kg to another planet more massive than Earth. Which statement is true about the prob
vivado [14]

Answer: The weight of the probe will increase.

8 0
3 years ago
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
3 years ago
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di
Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, u=20m/s

And the mass of the car is, m=1000 kg

The total distance covered by the car before stop, s=45m

And the final speed of the car is, u=0m/s

Now initial kinetic energy is,

KE_{i}=\frac{1}{2}mu^{2}

Substitute the value of u and m in the above equation, we get

KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J

Now final kinetic energy is,

KE_{f}=\frac{1}{2}mv^{2}

Substitute the value of v and m in the above equation, we get

KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN

Here, the force is negative because the force and acceleration in the opposite direction.

6 0
3 years ago
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