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Oksanka [162]
4 years ago
10

A battery and a resistor are wired into a circuit. The resistor dissipates 0.50 W. Now two batteries, each identical to the orig

inal one, are connected in series with the resistor. Part A What power does it dissipate?
Physics
2 answers:
laiz [17]4 years ago
7 0

Answer:

The resistor dissipates 1 W when two identical batteries are connected in series.

Explanation:

Power dissipated is directly proportional to the Voltage and Current.

Connected in series, the total voltage of two batteries is equal to the sum of the voltage of each one.

If I double the applied voltage, the dissipated power will be double.

P = V*I

morpeh [17]4 years ago
3 0

Answer:

  • <u>Power dissipation:</u>

The losses that occurs inside any system, or the amount of deficiency that happens in any system's power and it is mainly due to the resistive elements in the circuit or system called as the power dissipation.

As, the Power can be calculated by analyzing the amount of Potential difference,V and the flow of current,I between any two points that are under-observation. While we can write it as,

  • <u>Units:</u>

P=V.I,

And the unit is, Watt,W.

While we are provided by the power dissipation value, that is about 0.50 W.

And the two batteries,V are connected in series with the resistors,R. And we have connection given to us in series. So, the current,I distribution will be same all across the circuit.

<u></u>

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Answer:

W = 819152 J = 819.15 KJ

Explanation:

The work done by Juri can be given by the following formula:

W = FdCos\theta

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W = Work done = ?

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5 0
3 years ago
The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
FrozenT [24]

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

<em></em>

Explanation:

momentum of the electron before relativistic effect = M_{0} V

where M_{0} is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = M_{0}/ \sqrt{1 - \beta ^{2}  }

where

\beta = V/c

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> M_{0}V/ \sqrt{1 - \beta ^{2}  }

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75M_{0} V = M_{0}V/ \sqrt{1 - \beta ^{2}  }

the equation reduces to

1.75 = 1/ \sqrt{1 - \beta ^{2}  }

square both sides of the equation, we have

3.0625 = 1/(1 - \beta ^{2} )

3.0625 - 3.0625\beta ^{2} = 1

2.0625 = 3.0625\beta ^{2}

\beta ^{2} = 0.67

β = 0.819

substitute for  \beta = V/c

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>

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