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lutik1710 [3]
3 years ago
9

How does resistance affects current in a circuit???

Physics
2 answers:
zheka24 [161]3 years ago
4 0

Resistance decreases the amount of current in a circuit, because it opposes current.

Also, the relationship can be shown mathematically: Current=Voltage/Resistance

resistance is divided into the voltage, creating a smaller number which is the value for current.

kakasveta [241]3 years ago
3 0

Resistance decreases the flow of current in a circuit.

EXPLANATION:

The resistance of an electrical circuit is defined as the opposition to the flow of current in that particular circuit.

From Ohm's law, we know that current flowing  through a wire I = \frac{V}{R}

Here,V is the potential across the two ends of a conductor or wire, and R is the resistance of that wire.

Hence, more is the resistance, the less will be the current flow in the current.


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Answer:

Ff = u M g    frictional force where u = .62

1/2 M v^2    kinetic energy of water bottle at release

Ff * d = 1/2 M v^2 = u M g d   work to stop equals initial kinetic energy

v^2 = 2 u g d = 2 * .62 * 9.8 * 12 = 146 m^2 / s^2

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3 years ago
The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d
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Answer:

28.11m far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

  (1). \: \:\beta =10 log(\dfrac{I}{I_0})

The ratio of the intensities can be written as

$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $

\dfrac{I}{I_0} = \dfrac{A_0}{A}.

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A = 4\pi r^2

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A_0 = 4 \pi r_0^2,

\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2}  = \dfrac{r_0^2}{r^2}

meaning

\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.

Putting this into equation (1), we get:

\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

100dB=10 log(\dfrac{r_02}{(5m)^2})

10= log(\dfrac{r_0^2}{25})

by taking both sides to the exponent:  

10^{10}= \dfrac{r_0^2}{25}

r^2 = 25 *10^{10}\\r = 5 *10^5

Now equation (2) becomes

\beta = 10log(\dfrac{25*10^{10}}{r^2})

when the intensity level is 85 dB we have

85 = 10log(\dfrac{25*10^{10}}{r^2})

8.5 = log(\dfrac{25*10^{10}}{r^2})

take both sides to exponents and we get:

10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}

10^{8.5} =\dfrac{25*10^{10}}{r^2}

r^2 = \dfrac{25*10^{10}}{10^{8.5}}

\boxed{r = 28.11m}

Thus, 28.11m far from the speaker the intensity drops to 85 dB.

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