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Katyanochek1 [597]
2 years ago
7

A very long, straight wire has charge per unit length 3.20 x 10^-10 c>m. At what distance from the wire is the electric-field

magnitude equal to 2.50 nc?
Physics
1 answer:
vovangra [49]2 years ago
5 0

at 2.304×10⁹m the electric field magnitude equal to 2.50 nc.

electric field due to a long uniformly charged wire is given as,

E= 2kλ/r

where,

k is a constant

λ is uniform charge density

r is the distance from the wire

λ = 3.20 × 10⁻¹⁰C/m

k= 9×10⁹

therefore r = 2kλ/E

on substituting value we get,

r= 2.304×10⁹m

learn more about electric field due to a uniformly charged wire here:

brainly.com/question/28166144

#SPJ4

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Data:
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a. The speed of the tsunami at ocean depth of 8000 m is 1,019.66 km/h.

b. The speed of the tsunami at ocean depth of 3500 m is 660.75 km/h.

c. The speed of the tsunami at ocean depth of 70 m is 89.66 km/h.

d. The speed of the tsunami at ocean depth of 5 m is 30.625 km/h.

<h3>Speed of the tsunami at the given depths</h3>

The speed of the tsunami is calculated from method of extrapolation or interpolation as shown below;

<h3>Speed at 8000 meters</h3>

8000 m -------- ?

7000 m -------- 943 km/h

4000 m -------- 713 km/h

(8000 - 7000)/(7000 - 4000) = (? - 943)/(943 - 713)

0.3333 =  (? - 943)/230

230(0.3333) = ? - 943

76.66 = ? - 943

? = 1,019.66 km/h

<h3>Speed at 3500 m</h3>

4000 m ---------- 713 km/h

3500 m ---------- ?

2000 m ----------- 504 km/h

(4000 - 3500)/(4000 - 2000) = (713 - ?) / (713 - 504)

0.25 =  (713 - ?) /209

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? = 660.75 km/h

<h3>Speed at 70 m</h3>

200 m ------ 159 km/h

70 m --------- ?

50 m ------- 79 km/h

(200 - 70)/(200 - 50) = (159 - ?)/(159 - 79)

0.8667 = 159 - ? / 80

80(0.8667) = 159 - ?

69.336 = 159 - ?

? = 89.66 km/h

<h3>Speed at 5 meters</h3>

50 m ------- 79 km/h

10 m --------- 36 km/h

5 m ---------- ?

(50 - 5)/(50 - 10) = (79 - ?)/(79 - 36)

1.125 = 79 - ?/43

43(1.125) = 79 - ?

48.375 = 79 - ?

? = 30.625 km/h

Learn more about speed of tsunami here: brainly.com/question/23136161

#SPJ1

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