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klasskru [66]
3 years ago
7

1.D 2.A 3.A 4.im not sure please help 5.false One light bulb in a string of lights goes out. This causes all of the other lights

in the string to also go out. This is an example of a
A. resistor. B. parallel circuit. C. closed circuit. D. series circuit. 2.When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways?
A. An open circuit is being changed into a closed circuit.
B. A closed circuit is being changed into an open circuit.
C. A parallel circuit is being changed into a series circuit.
D. A series circuit is being changed into a parallel circuit.
3.Early telephone poles had wires that connected inside bell-shaped glass enclosures like the ones shown below.
several bell-shaped glass enclosures These glass enclosures helped to keep the electric current from moving outside the circuit of wires. The glass enclosures were used as
A. insulators. B. conductors. C. resistors. D. switches. 4.Use Ohm’s Law to determine the resistance in a circuit if the voltage is 12.0 volts and the current is 4.0 amps.
A. 8.0 ohms B. 48 ohms C. 3.0 ohms D. 12 ohms 5.In a closed circuit, the current only flows from the power source to an electrical device such as a lamp
A.TRUE B.FALSE
Physics
1 answer:
Amiraneli [1.4K]3 years ago
6 0
Welll ... if you read the question, you'll see what a jumbled mess you actually
posted, but I think I can pull enough out of it to give you some answers.

<span>One light bulb in a string of lights goes out. This causes all of the
other lights in the string to also go out. The entire string of lights
must be one series circuit.

2.  When a switch is turned from the off to the on position,
an open circuit is changed to a closed circuit.

3.Early telephone poles had wires that connected inside bell-shaped
glass enclosures like the ones shown below.  (nothing is shown below)
several bell-shaped glass enclosures
These glass enclosures helped to keep the electric current from moving
outside the circuit of wires. The glass enclosures were used as insulators.

4.Use Ohm’s Law to determine the resistance in a circuit if the
voltage is 12.0 volts and the current is 4.0 amps.

Ohms law:        Resistance = (voltage) / (current)

                                             = (12.0 v) / (4.0 amp)

                                             =      3 ohms .

5.  "In a closed circuit, the current only flows from the power source
to an electrical device such as a lamp."

This statement is false.  The current eventually has to get back
to the power source.  If it doesn't then there's no 'circuit', and
nothing works.  That's a big part of the reason why the plug
has 2 prongs on it, and the cord from the plug to the lamp
has 2 wires in it. </span>
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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

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Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

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