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kondaur [170]
3 years ago
11

What is the best defintion of efficency

Physics
2 answers:
asambeis [7]3 years ago
8 0

Answer: Efficiency signifies a peak level of performance that uses the least amount of inputs to achieve the highest amount of output.

Explanation: It minimizes the waste of resources such as physical materials, energy, and time while accomplishing the desired output.

Yuki888 [10]3 years ago
8 0

Answer:

The efficiency of a simple machine is defined as the ratio of useful work done by the machine ( output work) to the total work out into the machine ( input work).

Explanation:

<h3><u>Efficiency</u></h3>

If a machine overcomes a load ' L ' and the distance travelled by the load is 'Ld' , the work done by the load is L× LD. It is also called output work or useful work.

Therefore, \boxed{Output \: work \:  = L \:  \times  \: Ld}

Likewise, The effort applied to overcome the load is 'E' and the distance covered by effort is 'Ed' , the work done by effort is E × Ed. It is also called input work.

Therefore, \boxed{Input \: work = E  \times Ed}

The efficiency of a simple machine is defined as the ratio of output work to the input work .

Therefore, \boxed{Efficiency ( η)=  \frac{outpt \: work}{input \: work}  \times 100\%}

Efficiency is expressed in percentage. It is a ratio of two works. A machine is never 100% efficient. It is because no machine is friction free and due to friction, some of the input energy is wastes in the form of heat energy.

\mathrm{Hope \: I \: helped!}

\mathrm{Best \: regards!}

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A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric
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Answer:

(a) -6.76\times 10^{12}\ N/C

(b) -1.352\times 10^{13}\ N/C

(c) -7.2\times 10^{11}\ N/C

Explanation:

(a)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = 9\times 10^{9}\ Nm^2/C^2

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

E=(\frac{kQ}{R^3})r

Plug in the given values and solve for 'E'. This gives,

E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

(b)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

E=(\frac{kQ}{R^3})r

Plug in the given values and solve for 'E'. This gives,

E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

(c)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

E=\dfrac{kQ}{r^2}

Plug in the given values and solve for 'E'. This gives,

E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

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4 years ago
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