Question #3df20

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Question #3df20

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Answer 1 · 2017-01-20T01:38:09

See below.

Explanation:

The weight of a stationary object near the surface of the earth is equal the product of the mass of the object and the gravitational (free-fall) acceleration constant, $g$ $g$ . This is equal in magnitude to the force of gravity.

$W = {\to F}_{g} = m g$ $W=vecF_g=mg$

Where $W$ $W$ is the weight.

Weight is a force and can be given in newtons, where a newton is equivalent to one kilogram meter per square second:

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

If using newtons, the mass of the object should be in kilograms. The gravitational acceleration constant, $g$ $g$ , is equal to $9.8 \frac{m}{s^{2}}$ $9.8m/s^2$ . So, if we are calculating the weight of a stationary object on Earth we would have units:

$W = m g = k g \cdot (\frac{m}{s^{2}})$ $W=mg=kg*(m/s^2)$

$\Rightarrow \frac{k g m}{s^{2}}$ $=>(kgm)/s^2$

So, you should get units of newtons, $N$ $N$ .

Multiplying $k g \cdot \frac{N}{k g}$ $kg*N/(kg)$ would get you the correct units. The difference is that this expresses the units of acceleration as $\frac{N}{k g}$ $N/(kg)$ , which is equivalent to $\frac{\frac{k g m}{s^{2}}}{k g} = \frac{m}{s^{2}}$ $((kgm)/s^2)/(kg)=m/s^2$ . It is not incorrect.

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Question #3df20

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