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muminat
4 years ago
6

A positive charge moves in the direction of an electric field. Which of the following statements are true? Check all that apply.

Check all that apply. The potential energy associated with the charge increases. The amount of work done on the charge cannot be determined without additional information. The electric field does positive work on the charge. The electric field does not do any work on the charge. The potential energy associated with the charge decreases. The electric field does negative work on the charge. SubmitRequest Answer Provide Feedback Next
Physics
2 answers:
Sergio039 [100]4 years ago
5 0

Answer:

The electric field does positive work on the charge.

The potential energy associated with the charge decreases.

Explanation:

We know that force due to electric filed on a charge particle q given as

 F = q E          --------1

If the   positive charge is moving in the direction of electric filed it means that force due to electric field and displacement of particle due to motion is in the same direction.So the work done due to electric filed will be positive.

Work = Force . Displacement

W= F.d

F= W/d            ---------2

From equation 1 and 2

W/d = q E

W/ q = E/ d

We know that voltage difference ΔV

ΔV = E/ d

ΔV =W/ q

So the potential difference will be decrease because charge is positive.So the potential energy will be decreases.

jarptica [38.1K]4 years ago
3 0

Answer:

The potential energy associated with the charge decreases,

The electric field does positive work on the charge.

Explanation:

If a positive charge is moving in the direction of an electric field that it would normally move, the potential energy associated with the charge will decrease because the scenario is similar to what occurs in a constant gravitational field.

Also, when potential energy is lost (decreases), this means positive work is done on the charge.

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Substituting the given values then

0.400\times 11.5 = (0.400+75)v_c\\v_c=0.061007958\approx 0.061 m/s

(b)

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1=m_1v_2+m_2v_3

v_1 is the velocity of the ball , v_2 is the velocity of ball afterwards and v_3 is your speed, m_1 and m_2 are masses of the ball and person respectively. Since it bounces back, we give it a negative value hence

0.400\times 11.5= 0.4\times -7.8+75v_3\\v_3=0.102933333\approx 0.103 m/s

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A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
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Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

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d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

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     Emo = E_{mf}

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     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

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     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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