3 years ago

PLEASE HELP. Really important

Physics

1 answer:

Kruka [31]3 years ago

3 0

Explanation:

2) C would need the least effort, because the longer the effort distance, the least the effort applied.

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3 years ago

The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2

Artyom0805 [142]

Right, as you mentioned in the comments, you find $d$ by plugging in the different values of $t$ .

For $t=1\,\mathrm s$ , we have

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)$

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Similarly, for $t=2\,\mathrm s$ , you get

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)$

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3 years ago