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choli [55]
2 years ago
15

Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt

ery in each circuit is identical and each bulb (resistor) they add provides the same amount of resistance.
They notice that every time they added bulb 3 (R3) to Circuit A, the brightness of all the bulbs dimmed but when they added bulb 3 (R3), nothing changed in the brightness of the bulb.  Which explanation of this observation is most complete?

Question options:

The brightness of the bulbs is all about the total amount of voltage at each resistor: less voltage = dimmer bulbs.  Circuit A is a series circuit, so the voltage of the battery is divided across all of the resistors.  Circuit B is a parallel circuit so the voltage across each resistor in parallel is the same as the battery. 

The brightness of the bulbs is all about the total amount of current in the circuit: less current = dimmer bulbs.  Circuit A is a series circuit, the more resistors you add, the less total current you get.  Circuit B is a parallel circuit so the more resistors you add, the more total current you get. 

The brightness of the bulbs is all about the total amount of resistance in the circuit: more resistance = dimmer bulbs.  Circuit A is a series circuit, so every time you add a resistor the total resistance increases.  Circuit B is a parallel circuit, so every time you add a resistor the total resistance decreases.

The brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.


​

Physics
1 answer:
bija089 [108]2 years ago
4 0

The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and  current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A,  the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

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             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

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b) this part of the exercise we solve the speed of equation 1

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             v_{cm } = √(4/3 gh)

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             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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3 years ago
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