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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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Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here: 750 j electrical energy is
MAVERICK [17]

Here is the energy that is left after the quantity of energy is transformed: 750 j of electrical energy is changed into 400 j of kinetic or mechanical energy, which is then turned into 0.32 j of efficient energy.

To run the fan, electrical energy is utilized.

Here, under the specified circumstances, 750 J of electrical energy is utilized to operate the fan, which is transformed into 400 J of kinetic energy. As a result, 350 J of energy is wasted due to various frictional and resistive losses.

Therefore, we may conclude that only 400 J of the 750 J available energy is used to power the fan, with the remaining energy being wasted as a result of friction.

Additionally, we can state that this fan's effectiveness will be

n = Useful ÷ Total

n = 400 ÷ 750

n = 8 ÷ 25

n = 0.32

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4 0
1 year ago
How much work is done when an engine generates 400 Watts of power in 25 seconds?
IceJOKER [234]

Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

400 = workdone/25

workdone = 400 * 25

=10000 J

8 0
3 years ago
Read 2 more answers
A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist
Paha777 [63]

Answer:

Part a)

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

\frac{1}{2}mv^2 = mgh'

now we know that

I = \frac{2}{5}mr^2

\omega = \frac{v}{r}

so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

\frac{7}{10}mv^2 = mgh

\frac{1}{2}mv^2 = \frac{10}{14}mgh

so the height on the smooth side is given as

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

6 0
4 years ago
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True or False:
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This is a true statement
3 0
3 years ago
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Hi Guys.I was just wondering if given two specific heat capacities (In my case copper and water) do you add them both together o
timama [110]

Answer:

yes

Explanation:

using law of HC(heat capacity), which is

  • heat loss=heat gain
  • energy H=MCQ

Where M is mass of substance,C is specific heat capacity, and Q is temperature change

In case of two substance

  • the H = Mc*Cc*Q+Mw*Cw*Q(provided the initial and final temperature are given)

8 0
3 years ago
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