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Tasya [4]
3 years ago
7

PLEASE I NEED HELP

Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

K= 95.4 J

Explanation:

For this exercise we must use the conservation of mechanical energy.

We set a reference system on the floor.

Starting point. Higher

          Em₀ = U = m g h

Final point. Just before taking the floor

           Em_f = K = ½ m v²

energy is conserved because there is no friction

         Em₀ = Em_f

         mg h = K

         

The height is

  1.         h = y -y₀

                 h = 0- y₀

let's calculate

       K =  3.23 (-9.81)  (-3.01)          

       K= 95.4 J

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A car travels 90 meters north in 15 seconds. Then turns around and travels at 40 meters south in 5.0 seconds. what is the averag
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Average velocity  =  (displacement) / (total time)

Displacement = distance and direction between
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The car travels 90 meters north, then 40 meters south. 
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Displacement = 50 meters north
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The correct option is <u>D</u>.

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

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  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

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