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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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The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre
IrinaVladis [17]

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B) W = -320 KJ

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6 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
Ratling [72]

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

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Δv₃ = 5.282 m/s

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a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

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a₃ = a₂ + a₁

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F₂ = m x 0.3892...........(1)

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F₂/F₁

ratio = \dfrac{0.3892}{0.139}

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6 0
3 years ago
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