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Alekssandra [29.7K]
3 years ago
7

A diode has a power rating of 5W. If the diode voltage is 3.6V and the diode current is 1.75A, what is the power dissipation?​

Physics
1 answer:
son4ous [18]3 years ago
4 0

Answer:

Explanation:

Power = current * voltage

Power = 3.6 * 1.75

Power = 6.3 watts.

I hope this diode does not have to do this for very long. The power dissipation is above the diode's rated power -- a condition that won't last forever.

The power dissipation is 6.3 watts.

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The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif
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Hey again!

Ok..

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Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

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Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

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So

Q= Q' + Q"

Q= mc∆0 + ml

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when charges mutually repel and distribute themselves on the surface of conductors, what becomes of the electric field inside th
AleksAgata [21]

The charges align themselves so that the conductor's internal field is zero.

<h3>What occurs if a charged surface is in close proximity to a conducting surface?</h3>

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<h3>What takes place within a conductor?</h3>

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Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, the
ICE Princess25 [194]

Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by  \Delta L_{thick}and \Delta L_{thin}, respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e \frac{1}{2}   option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

               E = \frac{\sigma}{\epsilon}

 Where \sigma is the stress = \frac{Force}{Area}

             \epsilon is the strain = \frac{\Delta L}{L}

  Making Strain the subject

              \epsilon = \frac{\sigma}{E}

now in this question we are that the same tension was applied to both wires so

      \frac{\sigma}{E} would be constant

Hence

                 \frac{\Delta L}{L} = constant

for the two wire we have that

                  \frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}

      Looking at young modulus formula

                E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }

                    E * \frac{\Delta L }{L}  = \frac{F}{A}

                  A * \frac{\Delta L}{L}  = \frac{F}{E}

Now we are told that a comprehensive force is applied to the wire so for this question

                \frac{F}{E} is constant

And given that the length are the same

so  

     A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

  Assuming that

           A_2 = 2 A_1

So

       A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}

     \frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}

From the question the length are equal

      \Delta L_{thin} =2  \Delta L_{thick}

So  

       \frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}

Hence the ratio is less than 1

       

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