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Lelu [443]
4 years ago
7

Suppose a current-carrying wire has a cross-sectional area that gradually becomes smaller along the wire, so that the wire has t

he shape of a very long,truncated cone. How does the drift speed vary along the wire?
Physics
1 answer:
SIZIF [17.4K]4 years ago
3 0

Answer:

It slows down as the cross-section becomes smaller is the correct answer to this question.

Explanation:

  • That current is the same in all parts of the wire under steady-state conditions.
  • Thus the velocity of drift is approximately equal to the cross‐sectional region.
  • vd = I /nAq .
  • As the cross-section gets smaller it accelerates.
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3 years ago
A car travels 100 m while decelerating to 8 m/s in 5 s.<br> a) What was its initial speed?
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Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

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Answer:

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