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oksano4ka [1.4K]
3 years ago
10

When a can is crushed did it undergo a physical change

Physics
1 answer:
brilliants [131]3 years ago
4 0
Because even though  the object got crush and misshape it still has the same identity. the identity never change
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Find the value of currents through each branch
Irina-Kira [14]

Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

  (I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

__

It helps to be familiar with the formulas for resistors in series and parallel.

8 0
3 years ago
A mass of gas under constant pressure occupies a volume of 0.5 m3 at a temperature of 20°C. Using the formula for cubic expansio
Kryger [21]
No cubic expansion given
6 0
3 years ago
Which of the following is equivalent to 2.5 meters
blsea [12.9K]
B is the correct answer for sure bro
5 0
3 years ago
Read 2 more answers
Heat transfer from a hot surface to a cool surface in direct contact is called
navik [9.2K]
This is an example of conduction
8 0
3 years ago
A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1
kirza4 [7]

Answer:

\mu=41.5\textdegree

Explanation:

From the question we are told that:

Water index of refraction i_w=1.33

Glass index of refraction i_g=1.50

Generally the equation for Brewster's law is mathematically given by

 \theta=tan^{-1}(\frac{i_g}{i_w})

 \theta=tan^{-1}(\frac{1.50}{1.33})

 \theta=48.44 \textdegree

Therefore Angle of incident to plane  \mu (normal at 90 degree to the surface)

 \mu=90\textdegree-\theta

 \mu=90\textdegree-48.44\textdegree

 \mu=41.5\textdegree

8 0
2 years ago
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