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Gre4nikov [31]
3 years ago
14

A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon

d the edge hanging freely in air. how far beyond the edge of the wall can a 100kg woman walk before the plank began to rotate about the edge of the wall​
Physics
1 answer:
jek_recluse [69]3 years ago
7 0

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

            Στ = 0

            W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

           x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

            x₂ = \frac{M_1g  }{m_2 g} \ x_1

           

let's calculate

           x₂ = \frac{100}{75}  \ 1

           x₂ = 1.33 m

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4 0
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A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
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The change in potential energy is  \Delta  PE =  -  3.8*10^{-16} \ J

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  E =  950 \ N/C

      The  distance traveled by the electron is  x =  2.50 \ m

Generally the force on this electron is  mathematically represented as

     F =  qE

Where F is the force and  q is the charge on the electron which is  a constant value of  q =  1.60*10^{-19} \ C

    Thus  

      F  =  950  * 1.60 **10^{-19}

      F  = 1.52 *10^{-16} \ N

Generally the work energy theorem can be mathematically represented as

          W =  \Delta  KE

Where W is the workdone on the electron by the  Electric field and  \Delta  KE  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        W =  F* x  *cos(  \theta )

Where  \theta  =  0 ^o considering that the movement of the electron is along the x-axis  

        So

             \Delta  KE  =  F  * x  cos  (0)

substituting values

         \Delta  KE  =  1.52 *10^{-16}  * 2.50   cos  (0)

          \Delta  KE   =  3.8*10^{-16} J

Now From the law of energy conservation

       \Delta PE  =  -  \Delta  KE

Where \Delta  PE is the change  in  potential energy  

Thus  

        \Delta  PE =  -  3.8*10^{-16} \ J

               

7 0
3 years ago
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