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KatRina [158]
3 years ago
13

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

that must be applied at it's other hand to keep it in horizontal equilibrium is.​
Physics
1 answer:
solniwko [45]3 years ago
3 0

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

           30(30) = F(45)

                    F = 900/45 = 20 N

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Attempting to move a large rock useing a long lever will the work you do on the lever be greater than the same as or less than t
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3 years ago
An object of mass 20g taken at a height of 2m above the ground. Which type of energy is possessed by the object at this height ?
In-s [12.5K]
<h3><u>Given</u> </h3>
  • Mass of an object is \bf\red{20 g}
  • Height of the body is \bf\green{2 m}
<h3><u>To Find</u></h3>
  • Value of the energy

<h3><u>Solution</u></h3>

{\purple {\underline{\boxed{\sf{ Potential\: Energy \:=\: mgh}}}}}

Where,

  • \sf\red{m} = Mass
  • \sf\green{g} = Gravity
  • \sf\blue{h} = Height

Now, Converting gram to kg

\longrightarrow\: 1000g = 1kg

\longrightarrow\: 20g = \sf\dfrac{20}{1000}

\longrightarrow\: = 0.02 kg

According to the question

\sf\red{Mass \:= \:0.02 kg}

\sf\green{Gravitational \;force \:= \;10 m/s^{2}}

\sf\purple{Height\: =\: 2 m}

Putting these values

\to\: Potential Energy = mgh

\to\: Potential Energy = 0.02 × 10 × 2

\to\: Potential Energy = 0.2 × 2

\to\: Potential Energy = \bf\pink{0.4\:joules}

7 0
3 years ago
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