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mrs_skeptik [129]
3 years ago
15

A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

per unit volume is 8.49 × 1028 m-3. Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.
Physics
1 answer:
Karolina [17]3 years ago
3 0

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

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The gravitational potential energy will increase

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Near the Earth's surface, the GPE of an object is given by

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Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of th
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horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

x = vcos\theta * t

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solving above two equations we have

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now here we will plug in all data

0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}

0.9 = 150.65 - \frac{184150.2}{v^2}

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v = 35.1 m/s

<em>so the ball was hit with speed 35.1 m/s from the ground</em>

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