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muminat
3 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]3 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]3 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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What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total
Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

h_2 = \frac{4}{5}h

so the distance covered is given as

d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........

here we know that

h = 10 feet

d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)

d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})

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8 0
2 years ago
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8 0
2 years ago
Need help y’all ASAP please...physics
dolphi86 [110]

Answer:

t = 3/8 seconds

Explanation:

h=-16t^2 - 10t+6

h= 0 when it hits the ground

0=-16t^2 - 10t+6

factor out a -2

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divide by -2

0 = (8t^2 +5t -3)

factor

0=(8t-3) (t+1)

using the zero product property

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8t = 3         t= -1

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t cannot be negative  ( no negative time)

t = 3/8 seconds

3 0
2 years ago
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