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2 years ago

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In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????

in the wire?

1 answer:

34kurt2 years ago

3 0

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

$I = \frac{q}{t}$

<u>Where</u>:

q: is the electric charge transferred through the surface

t: is the time

The charge, q, is:

$q = n*e$

<u>Where</u>:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

$q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C$

Hence, the current in the wire is:

$I = \frac{126.88 C}{3.97 s} = 31.96 A$

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

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The frequency of oscillation is 2.153 Hz

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Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

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Substituting the values in the equation,

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Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

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This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
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Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

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Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

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