Physics, 27.11.2019 05:31 mkingston1705

Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. the first bead of mass m1=140gis placed on the top vertex. the second bead of mass m2=45g is placed on the left vertex. the third bead of mass m3=85g is placed on the right vertex.

(a) write a symbolic equation for the horizontal component of the center of mass relative to the left vertex of the triangle.

(b) find the horizontal component of the center of mass relative to the left vertex, in centimeters.

(c) write a symbolic equation for the vertical component of the center of mass relative to the base of the triangle.

(d) find the vertical component of the center of mass relative to the base of the triangle, in centimeters.

Answers: 1

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Answers

Answer from: scalderon2001

The center of mass is the point where the sum of the relative weighted

position of distributed masses m₁, m₂, and m₃ is zero.

The responses are as follows;

(a) The symbolic equation for $x_{cm}$ is $x_{cm} = \dfrac{ d \times \left(\dfrac{m_1}{2} + m_3 \right)}{m_1 + m_2 + m_3}$ (b) $x_{cm} = \underline{1.95\overline{185}}$ (c) The symbolic equation for $y_{cm}$ is $y_{cm} = \dfrac{m_1 \cdot d \cdot \dfrac{\sqrt{3} }{2}}{m_1 + m_2 + m_3}$ (d) $y_{cm}$ ≈ 1.53

Reasons:

The given masses and their location are;

Location of the masses = Thee vertices of the equilateral triangle

Side length of the equilateral triangle, d = 3.4 cm

m₁ = 140 g, location = Top vertex of the equilateral triangle

m₂ = 45 g, location = Left vertex of equilateral triangle

m₃ = 85 g, location = Right vertex of the equilateral triangle

Assumption;

The location of the left vertex of equilateral triangle = The origin (0, 0)

Therefore;

The location of the right vertex of equilateral triangle = (3.4, 0)

Location of the top vertex of the equilateral triangle = (3.4×cos(60°), 3.4×sin(60°))

Which gives;

Location of the top vertex of the equilateral triangle = (1.7, 1.7×√3)

(a) The symbolic equation for the horizontal component of the center of

mass relative to the left vertex is given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 + m_2 \cdot x_2 + m_3 \cdot x_3}{m_1 + m_2 + m_3}$

Where;

x₁ = x-coordinate of the top vertex = $\dfrac{d}{2}$

x₂ = x-coordinate of the left vertex = 0

x₃ = x-coordinate of the right vertex = d

Therefore, we have;

$x_{cm} = \dfrac{m_1 \cdot \dfrac{d}{2} + m_2 \times 0 + m_3 \cdot d}{m_1 + m_2 + m_3} = \dfrac{ d \times \left(\dfrac{m_1}{2} + m_3 \right)}{m_1 + m_2 + m_3}$

The symbolic equation for the horizontal component of the center of mass relative to the left vertex, in centimeters is $x_{cm} = \underline{ \dfrac{ d \times \left(\dfrac{m_1}{2} + m_3 \right)}{m_1 + m_2 + m_3}}$

(b) The horizontal component of the center of mass relative to the left

vertex is therefore;

$x_{cm} = \dfrac{ 3.4 \times \left(\dfrac{140}{2} + 85 \right)}{140 + 45 + 85} = \dfrac{527}{270} = 1.95\overline{185}$

$x_{cm} = \underline{1.95\overline{185}}$

relative to the left vertex is given as follows;

$y_{cm} = \dfrac{m_1 \cdot y_1 + m_2 \cdot y_2 + m_3 \cdot y_3}{m_1 + m_2 + m_3}$

Where;

y₁ = d × sin(60°) = $d \cdot \dfrac{\sqrt{3} }{2}$

y₂ = 0

y₃ = 0

Therefore;

$y_{cm} = \dfrac{m_1 \cdot d \cdot \dfrac{\sqrt{3} }{2}+ m_2 \times 0 + m_3 \times 0}{m_1 + m_2 + m_3} = \dfrac{m_1 \cdot d \cdot \dfrac{\sqrt{3} }{2}}{m_1 + m_2 + m_3}$

The symbolic equation for the vertical component of the center of mass

relative to the left vertex is therefore;

$y_{cm} = \underline{\dfrac{m_1 \cdot d \cdot \dfrac{\sqrt{3} }{2}}{m_1 + m_2 + m_3}}$

(d) The vertical component of the center of mass relative to the left vertex

is found by plugging in the values of the variables as follows;

$y_{cm} = \dfrac{140 \times 3.4 \times \dfrac{\sqrt{3} }{2}}{140 + 45 + 85} = \dfrac{119}{135} \cdot \sqrt{3}$

$y_{cm} = \underline{\dfrac{119}{135} \cdot \sqrt{3} \approx 1.53}$

$y_{cm}$ ≈ 1.53

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Answer from: stormhorn5108

Xcm = 1.95 cm and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

R cm = 1/M ∑ $m_{i}$ $r_{i}$

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

M = m₁ + m₂ + m₃

M = 140 + 45 + 85

M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

R₁ = d / 2 i ^ + d j ^

R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

R₂ = 0

Point 3. Right vertex

R₃ = d i ^

R₃ = 3.4 i ^ cm

a) The x component of the massage center

Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

Xcm = d / M (m₁ / 2 + m₃)

b) Let's write the mass center component x

Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

Xcm = 238/270

Xcm = 1.95 cm

c) let's find the component and center of mass

Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

Ycm = 1 / M (m₁ d + 0 + 0)

Ycm = m₁ / M d

d) let's calculate

Y cm = 1/270 (140 3.4 + 0 + 0)

Ycm = 1.76 cm

Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. the first bead

Answer from: Quest

The big glowy thing is called the sun it is a star in our galaxy and is over 1 million degrees

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.