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Eddi Din [679]
3 years ago
14

What is the difference between electric field and electrostatic field?

Physics
2 answers:
valentina_108 [34]3 years ago
7 0
Semantically? An electric field does not have to be static while in electrostatics it's generally assumed that the electric field does not change or only changes so slowly that its rate of change doesn't matter. That's very different in electrodynamics where electric and magnetic field changes are always coupled.

natta225 [31]3 years ago
3 0
The main difference<span> is that </span>Electrostatic field<span> is because of a stationary charge. And Induced </span>field<span> is because of a changing magnetic</span>field<span>. Since magnetic </span>field<span> and </span>electric field<span> are the same thing in different frames of reference.</span>
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A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​
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velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

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Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

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A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
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a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

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The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

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3 years ago
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