When does the potential energy of a swinging pendulum increase the most?

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

4 0

2 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

1 year ago

1. Johnny wants to know where the water line is in a dark well. He drops a penny into the well and counts until he hears the pen

777dan777 [17]

Answer:

3 feet down

Explanation:

i think

8 0

2 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

8 0

2 years ago

A piece of silicon sample has a resistivity of 0.1 ω.Cm. Its thickness is 100µm. The electron mobility is 1350cm 2 v -1sec-1. Wh

lawyer [7]

The answer is Rh = 135 cm^3 and B = 0.05185 wh/m^2

Explanation:

Resitivity of silicon = 0.1

thickness = 100um

so, I = ma

Required to find out concentration of electron , we know that

Rh = up

By putting in the values,

Rh = 1350 x 0.1

Rh = 135 cm^3

Now consider,

Rh = 1 / Rh.q

= 1 / Rh . q

= 1 / 135 x1.609 x10^-19

= 4.6037 x 10^16 / cm^3

Vh = BIRh / w

B = Vh w/ IRh

B = -70 x10^-6 x 100 x10^-6 / 1x 10^-3 x 135 x 10^-6

B = 0.05185 wh / m^2

5 0

3 years ago