Which of the following is the same size as Pluto?

xz_007 [3.2K]

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an <em>uniform</em> density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the <em>entire</em> system is found by applying definition of <em>weighted</em> averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$ , $W_{2}$ , $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

6 0

2 years ago

Un zapato de golf tiene 10 tacos cada uno con un área de 0.01 pulgadas en contacto con el piso suponga que el camino hay un inst

iren2701 [21]

Responder:

12374550,527 N / m²

Explicación:

Dado que:

Número de tacos = 10

Área de cada montante = 0.01 in²

Peso total de la persona = 180 libras

Usando la relación:

Presión = Fuerza / Área

La fuerza ejercida es el producto de la masa y la aceleración debido a la gravedad.

Masa = 180 libras = 81,47 kg

Aceleración por gravedad = 9,8 m / s²

Área total = (número de montantes * Área por montante) = (10 * 0.01) = 0.1 pulg²

Conversión a m²:

0,1 pulg² = 6,452 * 10 ^ -5 m²

Por lo tanto,

Presión = (81,47 * 9,8) / (6,452 * 10 ^ -5)

Presión = 798.406 / 0.00006452

Presión = 12374550.527 N / m²

6 0

3 years ago

S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr

Elina [12.6K]

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

#SPJ4

7 0

2 years ago

A diagram that shows thermal energy being released by objects is called a ______.

coldgirl [10]

The answer is a Thermogram.

I just took the test :)

3 0

3 years ago

Read 2 more answers

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>

5 0

3 years ago