# What is the weight (on earth) of a 13.5-gram object?

Mar 17, 2018

$0.1323$ Newtons $\left(N\right)$

$1$($N$)$=$1$\frac{k g \ast m}{{s}^{2}}$

#### Explanation:

The acceleration due to gravity, g, is $9.8$$\left(m\right)$/$\left({s}^{2}\right)$
Mass, m=$\left(13.5 g m\right)$**$\left(1 k g\right)$/$\left(1000 g m\right)$ =$0.0135$$\left(k g\right)$

The Weight force is $W = m g$$\left(N\right)$
Therefore
$W =$(0.0135$k g$)**(9.8$\frac{m}{s} ^ 2$) =$0.1323$$\left(N\right)$

Mar 17, 2018

$\approx 0.132 \setminus \text{N}$

#### Explanation:

Well, weight is expressed through the equation:

$W = m g$

where $m$ is the mass of the object in kilograms, $g$ is the gravitational acceleration constant, which is around $9.81 \setminus {\text{m/s}}^{2}$.

So, we first need to convert $13.5 \setminus \text{g}$ into $\text{kg}$. Recall that, $1 \setminus \text{kg"=1000 \ "g}$, so here, we got:

$13.5 \setminus \text{g"=0.0135 \ "kg}$

And so, weight of the object will be

$W = 0.0135 \setminus {\text{kg"*9.81 \ "m/s}}^{2}$

$= 0.132435 \setminus {\text{kg m/s}}^{2}$

But we know that $1 \setminus {\text{N"= 1 \ "kg m/s}}^{2}$, and so we got:

$= 0.132435 \setminus \text{N}$

$\approx 0.132 \setminus \text{N}$