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Furkat [3]
3 years ago
12

A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

ing the charge? A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in moving the charge? Work is required in moving the positive charge from point A to point B. No work is performed or required in moving the positive charge from point A to point B. Work is performed in moving the positive charge from point A to point B. Work is both performed and required in moving the charge from point A to point B. SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

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A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

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2 years ago
What kind of energy does a rubber band have when it is stretched?
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3 years ago
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Answer:

<h2>62.5 m/s</h2>

Explanation:

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d is the distance

t is the time

From the question we have

s =  \frac{125}{2}  = 62.5 \\

We have the final answer as

<h3>62.5 m/s</h3>

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