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Law Incorporation [45]
3 years ago
14

what are the common elements that make up series and parallel circuits , and how are circuits diagrammed?​

Physics
1 answer:
Ahat [919]3 years ago
3 0
<h2>Answer:</h2>

A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.

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10. What is momentum of the water bottle before the collision? (read the
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O ksm/s

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A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi
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Answer:

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0
3 years ago
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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of
mart [117]

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

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This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

5 0
3 years ago
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