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lana [24]
2 years ago
8

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 ?

Calculate the value of this energy?​​
Physics
1 answer:
In-s [12.5K]2 years ago
7 0
<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}

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Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

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  bottom center: I2+I3-I1 = 0 A

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Top left loop:

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Bottom left loop:

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In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

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