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Bezzdna [24]
3 years ago
7

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

iginally at 24°C, neglecting any changes in dimensions? Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is originally at 24°C, neglecting any changes in dimensions?
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

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Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

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The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

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\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

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4 years ago
What is the sum of 9260 and 3240?
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You can separate numbers out this way to make it easier to add them. 9260 can be broken down into 9000+200+60 while 3240 is 3000+200+40. You can then add these six numbers together.

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Then add your three partial results together to receive the final answer:

12000+400+100 = 12500
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algol13

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W = weight, m = mass, g = gravitational acceleration

Given values:

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Plug in and solve for W:

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