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Law Incorporation [45]
3 years ago
14

I need help with this question ASAP please :)

Physics
1 answer:
Nana76 [90]3 years ago
3 0

Answer: 1.67 x 10-7 N

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At twice the distance, the strength of the field is e/4

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F is inversely proportional to the squared distance from the point charge.

mathematically,

F = k(e/d²)

hence, if d is doubled,keeping other forms constant

F = e/4,

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When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length
monitta

Answer:

3.34\Omega

Explanation:

The resistance of a metal rod is given by

R=\frac{\rho L}{A}

where

\rho is the resistivity

L is the length of the rod

A is the cross-sectional area

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\rho(T)=\rho_0 (1+\alpha (T-T_0))

where in this case:

\rho_0 is the resistivity of silver at T_0=21.0^{\circ}C

\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1} is the temperature coefficient for silver

T=180.0^{\circ}C is the current temperature

Substituting,

\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0

The length of the rod changes as

L(T)=L_0 (1+\alpha_L(T-T_0))

where:

L_0 is the initial length at 21.0^{\circ}C

\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1} is the coefficient of linear expansion

Substituting,

L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0

The cross-sectional area of the rod changes as

A(T)=A_0(1+2\alpha_L(T-T_0))

So, substituting,

A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0

Therefore, if the initial resistance at 21.0°C is

R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega

Then the resistance at 180.0°C is:

R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega

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3 years ago
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