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1 year ago

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Three wires meet at a junction. wire 1 has a current of 0.40 aa into the junction. the current of wire 2 is 0.73 aa out of the j

unction. what is the magnitude of the current in wire 3?

1 answer:

Mnenie [13.5K]1 year ago

7 0

The magnitude of the current in wire 3 is (I₃)= 0.33A

<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>

To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,

I₁ + I₂ + I₃ = 0

Where we are given,

I₁ = current in wire 1

=0.40 A.

I₂ = current in wire 2

= -0.73 A.

We have to calculate the magnitude of the current in wire 3, I₃

Now we put the known values in above equation, we get,

I₁ + I₂ + I₃ = 0

Or, I₃ = -.(I₁ + I₂)

Or, I₃ = -.(0.40 - 0.73)

Or, I₃ = 0.33 A

From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A

Learn more about current:

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$8.3\,\%$ of that piece of ice would be above the freshwater. Assumptions:

the density of the ice is $\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}$ , and
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The volume of that chunk of ice can be split into two halves: volume above water $V(\text{above})$ , and volume under water $V(\text{under})$ . The mass of the whole chunk of ice would be:

$m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))$ .

Let $g$ be the acceleration due to gravity. The gravity on the entire chunk of ice would be

$\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

On the other hand, the size of buoyant force on an object is equal to the weight of the liquid that it displaces. That is: $F(\text{bouyancy}) = W(\text{water displaced})$ .

Recall that $V(\text{above})$ is the volume of the ice above the water, and $V(\text{under})$ is the volume of the ice under the water.

The mass of water displaced would be equal to:

$\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

The weight of that much water would be

$\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Apply the equation $F(\text{bouyancy}) = W(\text{water displaced})$ . The bouyant force on this chunk of ice would be equal to $\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Since the ice is floating, the forces on it need to be balanced. In other words, $\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

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