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Alex73 [517]
3 years ago
14

I NEED HELP PLEASE, THANKS! :)

Physics
2 answers:
brilliants [131]3 years ago
8 0

Answer:

P = VI = (IR)I = I2R

Explanation:

What the equation means is that if you double the current you end up with 4 times the power loss. It's like the area of carpet you need for a room - if you make the room twice as long and twice as wide you need 4x as much carpet. The physical explanation is that the voltage difference along a wire depends on the current - more current flowing with a resistance means more voltage (pressure of electricity if you like) is built up.

This extra voltage means more power. So if you double the current your would double the power, but you also double the voltage which doubles the power again = 4x as much power. P = VI = (IR)I = I2R

I hope this helps you out, if I'm wrong, just tell me.

Mumz [18]3 years ago
3 0

As I mentioned earlier, Ohm's law gives us the formula P = IV, where V is the voltage ( also known as the electrical potential difference ) and I is the current. It is confusing that P = I²R and P = IV are one in the same - so I want to go a bit deeper on that.

We have three formulas, P = IV, P = I²R, and P = V² / R. Each are considered the same. The two formulas P = I²R, and P = V² / R are derived from the statement that P = IV, under the condition V = IR. Substitute the value of V from this second condition V = IR into P = IV. You would get the following -

P = I( IR ),

P = I²R

That is how one can derive the formula P = I²R, and how P = IV and P = I²R are thought to be one in the same. If you would like, take a look at how to get the formula " P = V² / R, "

V = IR, P = IV

I = V / R, P = IV

P = ( V / R )V,

P = V² / R

<u><em>Hope that helps!</em></u>

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3 years ago
When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

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