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sweet [91]
3 years ago
14

Consider two wires made of copper. They have the same diameter but one of them has three times the length of the other. Choose t

he correct statement:
a. Both wires have the same resistivity
b. The long wire has one third of the resistivity of the short wire
c. None of the other answers is correct
d. The long wire has three times the resistivity of the short wire
Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

a. Both wires have the same resistivity

Explanation:

For the resistance of a wire , following formula holds good .

R = ρ l / S , R is resistance , l is length , S is cross sectional area and ρ is resistivity of the material that the wire is made of. Resistance is dependent on length and cross sectional area but resistivity does not depend upon length or cross sectional area . It only depends upon the type of material.

If we replace  copper wire with aluminium wire , then resistivity will change .

Hence , since the wire remains made of copper , resistivity will not change.

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Explanation:

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Two point charges of -7uC and 4uC are a distance of 20
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Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy \mathrm{EPE}.

\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r},

where

  • The coulomb's constant k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2},
  • q_1 and q_2 are the sizes of the two charges, and
  • r is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

  • q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C;
  • q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C;
  • Initial separation: \rm 20\; cm = 0.20\; cm;
  • Final separation: \rm 90\; cm = 0.90\; cm.

Apply Coulomb's law:

Initial potential energy:

\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}.

Final potential energy:

\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}.

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm  -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}.

8 0
3 years ago
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2 years ago
The drawing shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel a
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Answer:

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Explanation:

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ΣT=T_{N}+T_{f}+T_{W}

Since we are looking for when the wheel just starts to rise up N-> 0 so no torque due to normal force

T_{N}=0

The perpendicular lever arm for the F force is R-h

T_{f}=F*(r-h)

And the T of gravity according to the image

T_{W}=W*(\sqrt{r^2-(r-h)^2}

ΣT=0

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F*(r-h)+W*(\sqrt{r^2-(r-h)^2}=0

F=\frac{W*(\sqrt{r^2-(r-h)^2}}{r-h}

F=\frac{24.9 N*(\sqrt{0.336^2-(0.336-0.110)^2}}{(0.336-0.11)}

F=27.39N

4 0
3 years ago
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