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olganol [36]
2 years ago
8

*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a dis

tance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 300 N/C, pointing right.
What is the magnitude (in N) and direction of the electric force on the sphere?
magnitude_______N
Direction?
- toward the right
- toward the left
- the magnitude is zero


(b) What is the work (in J) done on the sphere by the electric force as it moves from A to B?
__________J


(c) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.)
PEB − PEA = ______J


(d) What is the potential difference (in V) between A and B?
VB − VA = ________V

Physics
1 answer:
pishuonlain [190]2 years ago
6 0

(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

(d) The potential difference between A and B is 56.7 V.

<h3>Electric force on the sphere</h3>

The electric force on the sphere is calculated as follows;

F = Eq

where;

  • E is electric field
  • q is the charge

F = 300 x (44 x 10⁻⁹)

F = 1.32 x 10⁻⁵ N

The direction of the force is towards the right.

<h3>Work done on the sphere</h3>

W = Fd

W = 1.32 x 10⁻⁵ N  x 0.189 m

W = 2.5 x 10⁻⁶ J

<h3>Change of the electric potential energy </h3>

The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.

<h3> Potential difference between A and B</h3>

VB − VA =  Ed

VB − VA =  300 N/C  x   0.189 m

VB − VA =  56.7 V

Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

The potential difference between A and B is 56.7 V.

Learn more about potential difference here: brainly.com/question/9060304

#SPJ1

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A 6 gram coin is which is initially at rest is dropped from the observation deck of a skyscraper 300 meters above the street bel
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Answer:

a) The work done by gravity on the coin as it falls is 17.653 joules.

b) The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

Explanation:

a) We must remind that situation with Earth-coin system must be represented by Principle of Energy Conservation and Work Energy Theorem. According to this latter, work done by gravity equals to the change in gravitational potential energy:

\Delta U_{g} = m\cdot g \cdot \Delta z (Eq. 1)

Where:

\Delta U_{g} - Change in gravitational potential energy, measured in joules.

m - Mass, measured in kilograms.

g - Gravitational accelerations, measured in meters per square second.

\Delta z - Height of the skyscraper, measured in meters.

If we know that m = 0.006\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta z = 300\,m, the work done gravity on the coin is:

\Delta U_{g} = (0.006\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (300\,m)

\Delta U_{g} = 17.653\,J

The work done by gravity on the coin as it falls is 17.653 joules.

b) By definition of translation kinetic energy, we get the following model:

K = \frac{1}{2}\cdot m\cdot v^{2}

Where:

K - Kinetic energy of the coin right before hitting the street, measured in joules.

v - Speed of the coin, measured in meters per second.

If we get that m = 0.006\,kg and v = 60\,\frac{m}{s}, the kinetic energy at this speed is:

K = \frac{1}{2}\cdot (0.006\,kg)\cdot \left(60\,\frac{m}{s} \right)^{2}

K = 10.8\,J

The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is obtained derived from Principle of Energy Conservation and Work-Energy Theorem:

W_{lost} = \Delta U_{g}-K (Eq. 2)

Where W_{lost} is the work lost due to air resistance, measured in joules.

If we know that \Delta U_{g} = 17.653\,J and K = 10.8\,J, the work lost due to air resistance is:

W_{lost} = 17.653\,J-10.8\,J

W_{lost} = 6.853\,J

The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance can be determined by applying definition of work, as air resistance force was antiparallel to the displacement of the coin. That is:

W_{lost} = F\cdot \Delta z

F = \frac{W_{lost}}{\Delta z} (Eq. 3)

Where F is the average force exerted on the coin due to air resistance, measured in newtons.

If we know that W_{lost} = 6.853\,J and \Delta z = 300\,m, then the average force exerted on the coin is:

F = \frac{6.853\,J}{300\,m}

F = 0.023\,N

The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

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3 years ago
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