The distance between the carbon atom and the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the carbon atom be at the origin $(0, 0)$ and the oxygen atom be placed at $(1.1, 0)$ on the $x$-axis.Given: $m_1 = 12 \ a.m.u.$ (carbon),$m_2

Vedclass · Accepted Answer

$0.63 \ \mathring{A}$ from the carbon atom

Vedclass · Answer

(C) Let the carbon atom be at the origin $(0, 0)$ and the oxygen atom be placed at $(1.1, 0)$ on the $x$-axis.Given: $m_1 = 12 \ a.m.u.$ (carbon),$m_2 = 16 \ a.m.u.$ (oxygen).Position vectors are $\vec{r}_1 = 0\hat{i}$ and $\vec{r}_2 = 1.1\hat{i}$.The center of mass $R_{cm}$ is given by:$R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$$R_{cm} = \frac{12(0) + 16(1.1)}{12 + 16}$$R_{cm} = \frac{17.6}{28} \ \mathring{A}$$R_{cm} \approx 0.63 \ \mathring{A}$ from the carbon atom.

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 \ \mathring{A}$. Given,mass of carbon atom is $12 \ a.m.u.$ and mass of oxygen atom is $16 \ a.m.u.$,calculate the position of the center of mass of the carbon monoxide molecule.

Similar Questions

The position of the centre of mass of a cube of uniform density will be at

The centre of mass of a thin rectangular plate (as shown in the figure) with sides of length $a$ and $b$,whose mass per unit area $(\sigma)$ varies as $\sigma = \frac{\sigma_0 x}{a b}$ (where $\sigma_0$ is a constant),would be . . . . . .

Define the centre of mass.

Find the position of the center of mass of a system of two particles of masses $m_1$ and $m_2$ separated by a distance $L$.

Give the location of the centre of mass of a $(i)$ sphere,$(ii)$ cylinder,$(iii)$ ring,and $(iv)$ cube,each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Vedclass Test Series

Exam Paper Generator

Online Exam Module