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Dimas [21]
3 years ago
5

What is the effective resistance of this dc circuit

Physics
1 answer:
Inessa05 [86]3 years ago
5 0

You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors.  I'm curious to know how you did that.

The three resistors at the left end of the diagram are  3Ω ,  2Ω , and  1Ω  all in series.  They behave like a single resistor of  (3+2+1) = 6Ω .

That  6Ω  resistor is in parallel with the  2Ω  drawn vertically in the middle of the diagram.  That combination acts like a single resistor of  1.5Ω  in that position.

Finally, we have that  1.5Ω  resistor in series with  1Ω  and  4Ω  .  That series combination behaves like a single resistor of  <em>6.5Ω</em>  across the battery V.  

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A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

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2 years ago
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume
Svetach [21]

Answer:

9.56\cdot 10^{-7} C

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

A=L^2=(0.19)^2=0.036m^2

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

E=3\cdot 10^6 N/C

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C

7 0
3 years ago
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