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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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Resistance = (voltage) / (current)

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<em>Resistance = 26 ohms</em>

4 0
4 years ago
Oxygen has 6 valence electrons. how many hydrogen atoms (who have 1 valence electrons) can bond with it? remember the octet rule
astra-53 [7]

Answer:

Oxygen starts with six valance electrons and ends with six valance electrons but after bonding has parts of the densities of 8 electrons

Explanation:

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The electron density of two of the valance electrons of Oxygen are shared with the Hydrogen atoms. The Hydrogen atoms is turn share part of the electron density of their single electron with Oxygen.

This means that there are parts of 8 electron densities around the Oxygen. This does not increase the number of valance electrons of Oxygen. (Note the sharing is very unfair, Oxygen gets more than its

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Explanation:

5 0
3 years ago
Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138
ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:  

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=\frac{1}{4}A_{o} is the final amount of the material

A_{o} is the initial amount of the material  

t is the time elapsed  

h=138 days is the half life of polonium-210

Knowing this, let's substitute the values and find t from (1):

\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}} (2)  

\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}} (3)  

\frac{1}{4}=2^{\frac{-t}{138 days}} (4)  

Applying natural logarithm in both sides:

ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}}) (5)  

-1.386=-\frac{t}{138days}ln(2) (6)  

Clearing t:

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3 years ago
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Answer:

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Explanation:

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A net force applied to a 15.0 kg box produced an acceleration of 4.2 m/s2. If the same net force was applied to a 10 kg box, wha
igor_vitrenko [27]

Answer:

6.3\ m/s^2

Explanation:

Given that,

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Force is given by :

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8 0
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