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Goshia [24]
3 years ago
10

Imagine that you have three circuit elements: a single bulb, a piece of wire, and a battery. You start by keeping these three el

ements separate and unconnected. You use a voltmeter to measure the voltage across each of the three elements. Then, you connect the bulb, wire, and battery together to form a complete circuit, so that the bulb lights up. You repeat the measurements you made before, using the voltmeter to measure the voltage across each element while it is in the complete circuit. For the questions below, assume that the battery and the wire are ideal.
a. Compare the voltage across the battery: a) when the battery is by itself vs b) when the battery is part of the complete circuit. Is the voltage across the battery different in these two situations, or is the voltage the same? Explain your answer.
b. Compare the voltage across the bulb: a) when the bulb is by itself vs b) when the bulb is part of the complete circuit. Is the voltage across the bulb different in these two situations, or is it the same? Explain.
c. Now compare the voltage across the piece of wire: a) when the wire is by itself vs b) when the wire is part of the complete circuit. Is the voltage across the piece of wire different in these two situations, or is it the same? Explain your answer.
d. Now consider only the situation in which the battery, wire, and bulb have been arranged to form a complete circuit in which the bulb lights up. For this situation, rank i) the voltage across the battery, ii) the voltage across the bulb, and iii) the voltage across the piece of wire, according to their absolute values, from greatest to smallest. If any are the same, or if any are zero, state that explicitly. Explain your reasoning.
Physics
1 answer:
EleoNora [17]3 years ago
8 0

Answer:

A)  two voltages are equal, B) the two voltages are different

C) voltage is the same V = 0, D)  V_{battery} = V_{Lamp} > V_{wire}

Explanation:

For this exercise, it is asked to build a circuit with the plug, the lamp and the battery, the circuit must be a series circuit

Before connecting the battery voltage is Vo as they indicate that it is ideal there are no losses, the voltages of the other elements are zero.

A) Voltages battery

a) single battery Vo

b) the battery is connected to the lamp this takes something current, but the electromotive force does not change therefore the voltage is Vo; This is because the battery's energy comes from a chemical reaction of the elements inside it.

in summary the two voltages are equal

B) voltage Lamp

a) when the bulb is single, its voltage is zero V = 0

since it has no energy

b) when in the circuit V = V₀ - V_r

where V_r is the voltage across the wire, due to its resistance, the energy for these voltages is given by the battery

in this case the two voltages are different

C) the voltage across the wire

a) unconnected V = 0

b) connected V = V₀ - V_L

where V_L is the voltage across the lamp due to its resistance

we can write the wire voltage

                V_r = i R

as they indicate that the wire is ideal, its resistance is zero R = 0, consequently

               V_r = 0

the voltage is the same V = 0

D) circuit voltage

 For this part we must write Ohm's law for this circuit

            V₀ = V_L + V_r

where V₀ is the battery voltage, V_L and V_r are the voltage of the lamp and the wire, respectively.

The voltage of each element is

                   V_L = i R_L

                   V_r = i R

This is a series circuit so the current is also constant

                 V₀ = i (R_L + R)

therefore the voltage in the battery is the highest V₀

the voltage across the wire if it is ideal is zero V_r = 0, if it is ideal its resistance is zero

therefore the voltage on the lamp is equal to the voltage of the battery

          V_{battery} = V_{Lamp} > V_{wire}

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y = y₀ + v₀ᵧ t + ½ aᵧt²

1 = 1 + 18 sin 52° t + ½ (-9.8) t²

0 = 14.2 t − 4.9 t²

0 = t (14.2 − 4.9 t)

t = 0 or 2.89

It takes 2.89 seconds to be caught.  The horizontal distance traveled in that time is:

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x = 0 + (18 cos 52°) (2.89) + ½ (0) (2.89)²

x = 32.1

Rounded to two significant figures, the ball travels 32 meters.

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The answer is "12682.267\ \ \frac{W}{m^2}"

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Calculating the amount of heat transfer.

q =m\timesC_p(T_{m,0}-T_{m,i})

  =0.0253 \times 4000 (75-25)\\\\=0.0253 \times 4000(50)\\\\=0.0253 \times 200,000 \\\\= 5060 \ W

Calculating the required value for heat flux:

q"=\frac{q}{A_s}

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3 years ago
How do ballerinas take advantage of conservation of momentum to turn?.
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Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
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Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

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Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
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  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

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