# Convert atomic unit of force to pond

Learn how to convert 1 atomic unit of force to pond step by step.

## Calculation Breakdown

Set up the equation
$$1.0\left(atomic \text{ } unit \text{ } of \text{ } force\right)={\color{rgb(20,165,174)} x}\left(pond\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(newton\right)$$
$$\text{Left side: 1.0 } \left(atomic \text{ } unit \text{ } of \text{ } force\right) = {\color{rgb(89,182,91)} 8.23872206 \times 10^{-8}\left(newton\right)} = {\color{rgb(89,182,91)} 8.23872206 \times 10^{-8}\left(N\right)}$$
$$\text{Right side: 1.0 } \left(pond\right) = {\color{rgb(125,164,120)} 9.80665 \times 10^{-3}\left(newton\right)} = {\color{rgb(125,164,120)} 9.80665 \times 10^{-3}\left(N\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(atomic \text{ } unit \text{ } of \text{ } force\right)={\color{rgb(20,165,174)} x}\left(pond\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 8.23872206 \times 10^{-8}} \times {\color{rgb(89,182,91)} \left(newton\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 9.80665 \times 10^{-3}}} \times {\color{rgb(125,164,120)} \left(newton\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 8.23872206 \times 10^{-8}} \cdot {\color{rgb(89,182,91)} \left(N\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 9.80665 \times 10^{-3}} \cdot {\color{rgb(125,164,120)} \left(N\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} 8.23872206 \times 10^{-8}} \cdot {\color{rgb(89,182,91)} \cancel{\left(N\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 9.80665 \times 10^{-3}} \times {\color{rgb(125,164,120)} \cancel{\left(N\right)}}$$
$$\text{Conversion Equation}$$
$$8.23872206 \times 10^{-8} = {\color{rgb(20,165,174)} x} \times 9.80665 \times 10^{-3}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$$8.23872206 \times {\color{rgb(255,204,153)} \cancelto{10^{-5}}{10^{-8}}} = {\color{rgb(20,165,174)} x} \times 9.80665 \times {\color{rgb(255,204,153)} \cancel{10^{-3}}}$$
$$\text{Simplify}$$
$$8.23872206 \times 10^{-5} = {\color{rgb(20,165,174)} x} \times 9.80665$$
Switch sides
$${\color{rgb(20,165,174)} x} \times 9.80665 = 8.23872206 \times 10^{-5}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{1.0}{9.80665}\right)$$
$${\color{rgb(20,165,174)} x} \times 9.80665 \times \dfrac{1.0}{9.80665} = 8.23872206 \times 10^{-5} \times \dfrac{1.0}{9.80665}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{9.80665}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{9.80665}}} = 8.23872206 \times 10^{-5} \times \dfrac{1.0}{9.80665}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{8.23872206 \times 10^{-5}}{9.80665}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.0000084012\approx8.4012 \times 10^{-6}$$
$$\text{Conversion Equation}$$
$$1.0\left(atomic \text{ } unit \text{ } of \text{ } force\right)\approx{\color{rgb(20,165,174)} 8.4012 \times 10^{-6}}\left(pond\right)$$