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olganol [36]
3 years ago
8

3 resistors of 3,4 and 12 ohm are there. How would you connect so as to get a resistance more than 5 oh, but less than 7 ohm sho

w using calculations and a diagram
Physics
1 answer:
DochEvi [55]3 years ago
4 0

Answer:

<em>Connecting the 4-ohm and 12-ohm in parallel and followed by the 3-ohm resistor in series. A scheme of the configuration is attached below. Please see the file attached below to know the diagram.</em>

Explanation:

There is the following solution that satisfies all requirements indicated on statement:

<em>Connecting the 4-ohm and 12-ohm in parallel and followed by the 3-ohm resistor in series. A scheme of the configuration is attached below.</em>

R = \frac{(12\,\Omega)\cdot (4\,\Omega)}{12\,\Omega+4\,\Omega}+3\,\Omega

R = 6\,\Omega

Which observes all design requirements.

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What kind of quantity is displacement?
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Read 2 more answers
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
3 years ago
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