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3 years ago

How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?

Physics

1 answer:

AleksandrR [38]3 years ago

7 0

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

WD= 45*6

WD=270Joues

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A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from

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3 years ago

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

$T = FR = 18 * 2.4 = 43.2 Nm$

According to Newton 2nd law, the angular acceleration would be

$\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2$

It starts from rest, then after 15s it would achieve a speed of

$\omega = \alpha t = 0.021 * 15 = 0.31 rad/s$

(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

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7 0

3 years ago

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

6 0

2 years ago

krek1111 [17]

Explanation:

the table and the wooden block

6 0

3 years ago

Read 2 more answers