KINDLYY FASTT A uniform metre rule of mass 100 g is pivoted at the 60 cm mark. At what point on the meter rule should a mass of 50 g be suspended for it to balance horizontally?

80 cm.

Explanation:

Since the metre rule is 1 m i.e 100 cm it means that the mass of the metre can be obtained at the 50 cm mark.

But the metre is pivoted at the 60 cm.

Please refer to the attached photo for details.

In the attached photo, y is the distance between the mark (that will balance the metre rule when a 50 g mass is hunged) and the pivot.

Thus, we can obtain the value of y as follow:

Anticlock wise moment = Clock wise moment

Anticlock wise moment = 100 × 10

Clock wise moment = y × 50

Anticlock wise moment = Clock wise moment

100 × 10 = y × 50

Divide both side by 50

y = (100 × 10) /50

y = 20 cm

Now, to obtain the mark, B that will balance the metre, we simply add 20 cm to 60 cm ie

B = 60 + 20

B = 80 cm

graphing motion: mastery test

209be

what is the instantaneous acceleration of

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0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

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