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Mama L [17]
3 years ago
14

A 1 kg rock is suspended by a massless string from one end of a meter stick at the 0 cm mark. What is the mass m suspended from

the meter stick at the 75 cm mark if the system is balanced?
Physics
1 answer:
enot [183]3 years ago
8 0

Answer:

2 kg

Explanation:

Note: For the meter stick to be balanced,

Sum of clock wise moment must be equal to sum of anti clock wise moment

Wd = W'd' ................ Equation 1

Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.

make W' the subject of the equation

W' = Wd/d'............... Equation 2

Taking our moment about the support,

Given: W = mg =  1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm

Substitute into equation 2

W' = 9.8(50)/25

W' = 19.6 N.

But,

m = W'/g

m = 19.6/9.8

m = 2 kg.

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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
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