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Sonbull [250]
2 years ago
8

A lightweight plastic rod has a mass of 2.0 kg attached to one end and a mass of 2.5 kg attached to the other end. The rod has a

length of 1.20 m. How far from the 2.0-kg mass should a string be attached to balance the rod
Physics
1 answer:
ipn [44]2 years ago
5 0

The distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.


<h3>What is distance?</h3>

Distance can be defined as the horizontal length between two points.

To calculate the distance at which the string must be attached to balance the rod, we use the formula below.

Formula:

  • m₁gx = m₂g(1.2-x)................. Equation 1

From the question,

Given

  • m₁ = 2.0 kg
  • m₂ = 2.5 kg
  • x = Distance of the string from the 2.0 kg mass
  • g = Acceleration due to gravity.= 9.8 m/s²

Substitute these values into equation 1

  • 2(9.8)(x) = 2.5(9.8)(1.2-x)

Solve for x

  • 19.6x = 29.4-24.5x

Collect like terms

  • 19.6x+24.5x = 29.4
  • 44.1x = 29.4
  • x = 29.4/44.1
  • x = 0.67 m

Hence, the distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.

Learn more about distance here: brainly.com/question/17273444

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Answer:

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Explanation:

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Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potentia
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I think The choose (B)

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A backpacker walks for 3 days in the mountains and covers 16 km. How fast was he walking in km/hr?
Fudgin [204]

Answer:

0.22 km/h

Explanation:

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16/72 = 0.222222222

round to 0.22 km/h

For future reference, Distance/Time= Speed

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3 years ago
What is your velocity if it takes you 40 seconds to walk around a circle of 50meters 3 times?​
Shtirlitz [24]

Answer:

0 m/s

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8 0
3 years ago
Read 2 more answers
A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
Lapatulllka [165]

Answer:

Explanation:

The mass of that science book...wow. In pounds that would be 35.2! Yikes!

Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is

v^2=v_0^2+2aΔx  where

v is the final velocity, our unknown,

v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,

a is the pull of gravity which is always -9.8 m/s/s, and

Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:

v^2=0^2+2(-9.8)(-120) so

v=\sqrt{2(-9.8)(-120)} and

v = 48 m/s

As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:

12^2=0^2+2(-9.8)Δx and

144 = -19.6Δx so

Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:

120 - 7.3 = 112.7 m off the ground.

6 0
3 years ago
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