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swat32
3 years ago
14

Two long, parallel wires carry currents of different magnitudes. if the amount of current in each wire is doubled, what happens

to the magnitude of the force between the wires
Physics
1 answer:
Mrac [35]3 years ago
4 0
The force per unit of length between two wires carrying current is
\frac{F}{L}= \frac{\mu_0 I_1 I_2}{2 \pi r}
where I1 and I2 are the currents in the two wires, while r is the distance between them.

We can see from the formula that the force is proportional to the product between I1 and I2: F \sim I_1 I_2
so, if we double both I1 and I2, we get a factor 4:
F' \sim (2I_1 )(2I_2)=4 I_1 I_2 =4 F
so, the force between the wires will be 4 times the original value.
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nexus9112 [7]
Carbon is the answer to the problem
4 0
3 years ago
If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk
Westkost [7]

Answer:

1.843 x 10^-5 C  

Explanation:

<u><em>Givens:   </em></u>

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.  

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation  

E = (Q/A)/∈o                                (1)

Where,

Q: total charge on the disk.

A: the area of the disk.  

<u><em>Calculations:  </em></u>

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold  

Q = EA∈o

   = (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)  

  = 1.843 x 10^-5 C  

note:

calculations maybe wrong but method is correct

8 0
3 years ago
What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?
iren2701 [21]

Answer:

  • 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

V(r) = \frac{4}{3} \ \pi \ r^3

If we got an uncertainty \Delta r the formula for the uncertainty of V is:

\Delta V(r) = \sqrt{  (\frac{dV}{dr} \Delta r)^2  }

We can calculate this uncertainty, first we obtain the derivative:

\frac{dV}{dr}  = 3 * \frac{4}{3} \ \pi \ r^2

\frac{dV}{dr}  = 4 \ \pi \ r^2

And using it in the formula:

\Delta V(r) = \sqrt{  (4 \ \pi \ r^2\Delta r)^2  }

\Delta V(r) = \sqrt{  4^2 \ \pi^2 \ r^4 \Delta r^2  }

\Delta V(r) =  4 \  \pi \ r^2 \Delta r

The relative uncertainty is:

\frac{\Delta V(r)}{V(r)}

\frac{ 4 \  \pi \ r^2 \Delta r  }{ \frac{4}{3} \ \pi \ r^3}

\frac{ 3  \Delta r  }{  r}

Using the values for the problem:

\frac{ 3 * 0.09 m  }{  5.66 m} = 0.0477

This is, a percent uncertainty of 4.77 %

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2 years ago
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What is the volume of this bubble when it reaches the surface?
steposvetlana [31]

Answer:

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8 0
2 years ago
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