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Vaselesa [24]
3 years ago
10

A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be tru

e for this case? A) The negative charge performs work in moving from point A to point B. B) Work is required to move the negative charge from point A to point B. C) No work is required to move the negative charge from point A to point B. D) The work done on the charge depends on the distance between A and B. E) Work is done in moving the negative charge from point A to point B.
Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

C) No work is required to move the negative charge from point A to point B.

Explanation:

An equipotential surface is defined as a surface connecting all the points at the same potential.

Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.

The work done when moving a charge is given by

W=q\Delta V

where

q is the charge

\Delta V is the potential difference between the initial and final point of motion of the charge

However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

\Delta V=0

And so, the work done is also zero.

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An aircraft flies at sea level at a speed of 220 m/s. What is the highest pressure that can be acting on the surface of the airc
goldenfox [79]

Answer:

An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

Explanation:

Applying Bernoulli's equation, we determine the highest pressure on the aircraft.

P = \frac{1}{2} \rho V^2

where;

P is the highest pressure on the aircraft

\rho is the density of air = 1.204 kg/m³ at sea level temperature.

V is the velocity of the aircraft = 220 m/s

P = 0.5*1.204*(220)² = 29136.8 N/m²

Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

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3 years ago
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A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnit
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Answer:

2

Explanation:

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A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

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3 years ago
A 10.3 kg block of ice slides without friction down a long track. The start of the track is 4.2 m higher than the end of the tra
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Hello

1) Since there is no friction between the ice and the track, there is no loss of energy in the motion, so we can apply the law of conservation of energy.
The total energy E (sum of potential energy P and kinetic energy K) must be conserved:
E=P+K  

2) At the beginning of the motion, the total energy of the object is just potential energy:
E_1=P=mgh 
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3) At the end of the motion, this potential energy has converted into kinetic energy, and so the total energy at this point is 
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where m is the mass and v is the final velocity of the object.

4) We said that the total energy must be conserved, therefore we can write
E_1 = E_2
and so:
mgh= \frac{1}{2}mv^2
from which we can find v, the velocity:
v= \sqrt{2gh}= \sqrt{2\cdot9.81~m/s^2 \cdot 4.2~m}=9.08~m/s
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