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timofeeve [1]
3 years ago
15

We can talk about the amountofpotential energy per unit chargeas a function of locationâwe use the term voltageto describe this.

A voltage difference between two locations results from the presence of an electric field. The more work the field can do in moving thecharge from point A to point B, the higher the voltageof point A is relative to point B.
Required:
a. Describe and draw the electric field between two large, oppositely charged plates.
b. Electric potentialis completely analogous to an object being dropped off of a building. The higher the building, the more potential energy the object has. Make a sketch of the ground and a building anddraw lines that indicate constantpotential energyof an object.
c. On your sketch for part a, draw lines of constant potential (these are called equipotentiallines). It may help to imagine placing a positive charge between the plates. It will accelerate toward the negative plate and run into it. Regardless of where you place the charge on an equipotential line, itâll hit the negative plate with the same speed.

Physics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

potential lines that are perpendicular to the electric field, they must be parallel to the plates

Explanation:

A and C) for this drawing, as the electric field lines go from the positive plate to the negative plate and are horizontal to the potential lines that are perpendicular to the field, they must be parallel to the plates, see attached part a

B) in this case the ground has an energy of zero and the lines are parallel to the ground, see attached part b

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The rate at which light energy is radiated from a source is measured in which of the following units?
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g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
3 years ago
A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio
Nady [450]

Answer:

The speed of player is given by

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Explanation:

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T=\frac{2vsin\alpha }{g}    (i)

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Now the person must also reach the impact point of ball in the same time as above.

Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

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Now the distance the player must cover is given by

D= R-d

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D=\frac{2v^{2}sin2\alpha-gd}{g}  (ii)

Now the average speed of player is given by

V=\frac{D}{T}   (iii)

Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

V=\frac{\frac{2v^{2} sin2\alpha-gd }{g}}{\frac{2vsin\alpha }{g} }

V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}

3 0
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