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3 years ago

6

Charges q, q, and – q are placed on the x-axis at x = 0, x = 4 m, and x = 6 m, respectively. At which of the following points do

es the electric field have the greatest magnitude?

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

can you show a graph but if not i believe the answer is x=6m

Explanation:

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In a chemical reaction, one element replaces another element in a compound to form a new substance. Which of the following best

Andreyy89

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single replacement

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3 years ago

Tungsten has a temperature coefficient of resistivity of 0.0045 (c°)-1. a tungsten wire is connected to a source of constant vol

Murljashka [212]

Answer:

$131.1^{\circ}C$

Explanation:

The power delivered in the wire is given by:

$P=\frac{V^2}{R}$

where V is the voltage of the battery and R is the resistance of the wire.

Since the voltage of the battery is constant, we can rewrite this equation as follows:

$V^2 = PR=const.$ (1)

At the beginning, the initial resistance is $R_0$ , and the power delivered is $P_0$ . Later, when the temperature increases, the power becomes $P_1 = \frac{2}{3}P_0$ , and the new resistance is $R_1$ . Using (1), we can write

$P_0 R_0 = \frac{2}{3}P_0 R_1\\R_1 = \frac{3}{2}\frac{P_0 R_0}{P_0}=\frac{3}{2}R_0$ (2)

So, the new resistance must be 3/2 of the initial resistance.

We know that the resistance increases linearly with the temperature, as

$R_1 = R_0 (1+\alpha \Delta T)$

where

$\alpha = 0.0045 ^{\circ}C^{-1}$ is the temperature coefficient

$\Delta T$ is the change in temperature

Using (2), we can rewrite this equation as

$\frac{3}{2}R_0 = R_0(1+ \alpha \Delta T)$

and we find:

$\frac{3}{2}=1+\alpha \Delta T\\\Delta T=\frac{\frac{3}{2}-1}{\alpha}=111.1 ^{\circ}$

So, the new temperature of the wire must be

$T_f = 21^{\circ}+111.1^{\circ}=132.1^{\circ}$

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4 years ago