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nasty-shy [4]
3 years ago
7

Please help me with this physics problem​

Physics
1 answer:
Vesna [10]3 years ago
4 0

Answer:

idk

Explanation:

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A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 m from the b
EleoNora [17]

Answer:

  • 0.317 s

Explanation:

distance of fan A = 18.3 m

distance of fan B = 127 m

speed of sound (s) = 343 m/s

What is the time difference between hearing the sound at the two locations?

time (T) = distance / speed

  • time for sound to reach fan A = 18.3 / 343 = 0.053 s
  • time it takes for sound to reach fan B = 127 / 343 = 0.370 s
  • time difference = 0.370 - 0.053 = 0.317 s
5 0
3 years ago
Which of the following is a zone found in the open ocean
Svetlanka [38]
What are your answer choices?

5 0
3 years ago
Read 2 more answers
The density of iron is 7.87 g/cm3. If an iron cannonball has a volume of 105 cm3
Reptile [31]

the answer is 0.284 lb/in3

6 0
2 years ago
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
bekas [8.4K]

Answer:

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

Explanation:

This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.

The speed of sound in solid soil is an average of the speed of its constituent media, giving values ​​between

 wood      3900 m / s

 concrete 4000 m / s

 fabrics     1540 m / s

 earth       5000 m / s wave S

 ground    7000 m / s P wave

 

we can see that the speed on solid earth is an order of magnitude greater than in air.

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

From the initial information, the wave going through the ground should arrive first.

3 0
3 years ago
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