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ziro4ka [17]
4 years ago
11

Determine the power of 60 joules of work done in four seconds

Physics
1 answer:
zheka24 [161]4 years ago
7 0
Power = (energy) / (time)

           =   (60 joules) / (4 seconds)

           =         15 joules/second

           =          15 watts  
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Which formation is one feature of karst topography?
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Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves. It has also been documented for more weathering- resistant rocks, such as quartzite, given the right conditions.
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3 years ago
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In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
4 years ago
Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?
Ray Of Light [21]

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

           P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

            i = i₁ + i₂

if the two light bulbs are the same

           i = 2 i₁

           i₁ = i / 2

so the power of each bulb is is

           P = i₁² R

           P = R i² / 4

           P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

3 0
3 years ago
What is four possible energy sources for a circuit
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