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3 years ago

What do you call the process of finding the magnitude of fy

Physics

2 answers:

vlada-n [284]3 years ago

4 0

Answer:

The process of finding the magnitude of Fy is called vector resolution.

Explanation:

Given a vector with a magnitude equals to F which forms an angle equals to θ with the x-axis, then it can be decomposed into two smaller vectors (Fx and Fy) as follows:

Fx = F*cosθ (placed on the x-axis)

Fy = F*sinθ (placed on the y-axis)

This process is called vector resolution.

MAVERICK [17]3 years ago

3 0

Answer - vector resolution

Any vector suppose f if lies at some angle, then

fy = F x cos or sin (theta)

and fx = F x cos or sin (theta)

Process of finding the value of component of any vector quantity is called Vector Resolution.

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3 years ago

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proport

netineya [11]

Answer:

$[F]=[MLT^{-2}]$

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

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We need to find the dimensions of force. The dimension of force m and a are as follows :

$[m]=[M]$

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3 years ago

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3 years ago

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Ronch [10]

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Mirages happen when the ground is very hot and the air is cool.

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1 year ago

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Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su

dybincka [34]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ( $v$ ), en metros por segundo:

$v = \sqrt{\frac{K}{\rho} }$ (1)

Donde:

$K$ - Módulo de compresibilidad, en newtons por metro cuadrado.
$\rho$ - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que $\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}$ y $K = 2,16\times 10^{9}\,\frac{N}{m^{2}}$ , entonces la velocidad de las ondas sonoras es:

$v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }$

$v\approx 1469,694\,\frac{m}{s}$

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ( $\lambda$ ), en metros, mediante la siguiente fórmula:

$\lambda = \frac{v}{f}$ (2)

Donde $f$ es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que $v\approx 1469,694\,\frac{m}{s}$ y $f = 1000\,hz$ , entonces la longitud de onda de las ondas sonoras es:

$\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}$

$\lambda = 1,470\,m$

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

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2 years ago