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kifflom [539]
4 years ago
15

You short-circuit a 18 volt battery by connecting a short wire from one end of the battery to the other end. if the current in t

he short circuit is measured to be 19 amperes, what is the internal resistance of the battery\
Physics
1 answer:
mina [271]4 years ago
3 0
The relationship between the voltage and the current in the circuit is:
\Delta V = RI
where \Delta V is the voltage difference, R the resistance of the circuit (and therefore, the resistance of the battery) and I the current flowing in the circuit. By using I=19 A and \Delta V = 18 V, we can find the internal resistance of the battery:
R= \frac{\Delta V}{I}= \frac{18 V}{19 A}=0.95 \Omega
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<u>Answer</u>

5) b-c

6)    a-b and

       e-f

7) f-g

9) a-b = 0 m/s

    c-d = 0.6667 m/s

    e-f = 0 m/s

    f-g = -3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

<u>Explanation</u>

Answer

5) b-c

In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.

6) a-b and e-f

At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.

7) f-g

At this section the slope is negative meaning the cart is moving back to where it came from.

9) a-b = 0 m/s

At a-b the cart is not moving. So the velocity is zero.

<u>     c-d = 0.66667 m/s</u>

Velocity = distance / time

               =(50-40)/(40-25)

                = 10/15

                 = 0.6667  m/s

   <u> e-f = 0 m/s</u>

At e-f the cart is not moving. So the velocity is zero.

 <u>   f-g = -3 m/s</u>

Velocity = distance / time

               = (60-30)/(65-75)

                = 30/-10

                = - 3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

     

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