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STatiana [176]
3 years ago
15

5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.

(1 pt) Compare the currents in the two segments. That is, is I2 greater than, less than, or equal to I1? Explain. b. (1 pt) Compare the current densities J1 and J2 in the two segments. c. (1 pt) Compare the electric field strengths E1 and E2 in the two segments. d. (1 pt) Compare the drift speeds vd1 and vd2 in the two segments.
Physics
1 answer:
Volgvan3 years ago
7 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same  i.e. I1 = I2  and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment  therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

 note : electric field strength ∝ current density

since current density of first segment is > current density of second segment  and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because  ; vd ∝ current density

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maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

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   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

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4 0
4 years ago
Using this information...
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v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)

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Answer:

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There's not enough information to find an answer.

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