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faltersainse [42]
3 years ago
10

Equipotentials are lines along which Select one: a. the electric field is constant in magnitude and direction. b. the electric c

harge is constant in magnitude and direction. c. maximum work against electrical forces is required to move a charge at constant speed. d. a charge may be moved at constant speed without work against electrical forces. e. charges move by themselves.
Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

option D is correct

Explanation:

It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along an equipotential, since ΔV = 0. Thus the work is :

                                W = −ΔPE = −qΔV = 0.

Work is zero if force is perpendicular to motion. Force is in the same direction as E, so that motion along an equipotential must be perpendicular to E. More precisely, work is related to the electric field by:

                                 W = Fd cos θ = qEd cos θ = 0.

- The change in kinetic energy Δ K.E by conservation should be:

                                 Δ K.E = W

Since, W = 0:

                                Δ K.E = 0

- If change in kinetic energy is zero it means that charge moves at a constant speed. Hence, option D is correct.

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A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 45.0 deg
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a. 139.748J

b.3.486m/s

c.zero

Explanation:

a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°

Potential energy during release is calculated as:PE=mgh

#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.

Find the height when the child is pulled back:

cos 45\textdegree=y/2.10\\\\y=1.48m

#therefore,vertical difference is 2.1m-1.48m=0.62m

\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J

#Hence the potential energy during release is 139.748J

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At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:

KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s

#Hence the velocity at the bottom of the swing is 3.486m/s

c. Work is calculated as the product of force by distance.

From a, b above we have mass as 23.0kg .

-since the distance of the ropes remained constant the change in distance is zero:

W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0

Therefore the work in the ropes is 0,zero.

8 0
3 years ago
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