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xz_007 [3.2K]
3 years ago
11

A positive charge is moved from one point to another point along an equipotential surface. The work required to move the charge

_________.a. depends on the sign of the potential.b. is zero. c. is positive. d. is negative.e. depends on the magnitude of the potential.
Physics
1 answer:
Strike441 [17]3 years ago
5 0

Answer:zero

Explanation:

Work done in moving a charge is zero in equipotential  surface because potential of each point is same .

When there is Potential gradient then some particle experiences a force while moving but in equipotential surface there is no potential gradient , that is why work done in moving a charge is zero.

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What is the difference between an induced and a permanent magnet?
MissTica
INDUCTION MOTOR:-

Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control


Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation

Power density:-Induction produced by squirrel cage rotor inherently limits power density

Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos

Cost:-Relatively modest initial cost; higher operating costs

PERMANENT MAGNET MORTOR:-

speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range

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power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.

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8 0
3 years ago
A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest
Ludmilka [50]
(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is, mgh=\frac{mv^2}{2} or 2gh=v^2 (the mass doesn't even matter). The result is \sqrt{2gh}, so only the height matters really. It is almost 9 (it is \sqrt{80}=4\sqrt{5}).
6 0
4 years ago
Use Newton’s method to solve the equation0 =12+14x2−xsin(x)−12cos(2x),withx0=π2.Iterate using Newton’s method until an accuracy
Charra [1.4K]

Answer:

given function is

f(x)=12+14x^2-xsin(x)-12cos(2x)

x_{0} =\pi ^2

formula for newton's method is

x_{n+1} =x_{n} -f(x_{n})/f'(x_{n} )

so derivative of function is

f'(x)=28x-sin(x)-xcos(x)+24sin(2x)

now put values and solve

or you can also use MATLAB code to solve

i.e

function p= newton(x)

e=0.001;

for i=1:100

    if abs(d(x))>e

       if abs(k(x))>0

  xm=x-(k(x)/d(x));

  x=xm;

       else

       end

        break;

   end

end

    disp(x)

 disp(k(x))

return;

   

7 0
3 years ago
What kind of reaction (endothermic or exothermic)
baherus [9]
Exothermic is the right kind of reaction.
4 0
4 years ago
A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 3.10 T dir
aliya0001 [1]

Explanation:

Given that,

Area enclosed by a brass bracelet, A=0.005\ m^2

Initial magnetic field, B_i=3.1\ T

The electrical resistance around the circumference of the bracelet is, R = 0.02 ohms

Final magnetic field, B_f=0.93\ T

Time, t=16\ ms=16\times 10^{-3}\ s

The expression for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_f-B_i}{t}

\epsilon=-0.005 \times \dfrac{0.93-3.1}{16\times 10^{-3}}

\epsilon=0.678\ volts

So, the induced emf in the bracelet is 0.678 volts.

Using ohm's law to find the induced current as :

V = IR

I=\dfrac{V}{R}

I=\dfrac{0.678}{0.02}

I = 33.9 A

or

I = 34 A

So, the induced current in the bracelet is 34 A. Hence, this is the required solution.

5 0
3 years ago
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