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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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To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

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At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

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$\omega =$ Angular velocity

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Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

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Replacing at the previous equation we have that the angular velocity is

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$\omega = \frac{ 16.67}{0.25}$

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Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

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(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

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Average velocity = $\frac{Speed}{time}$

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Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

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(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

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(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

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Now substituting 4 in 3 we get

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$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

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