It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applications, from navigation to power generation. One of the first such experiments involving this technique was an August 1992 space shuttle flight, but the tether failed and only only 250 m of the conducting wire could be let out. A 40.0 V motional emf was generated in the Earth's 5.0 x 10-T field, while the shuttle and tether were moving at 7.80 x 10 m/s. What was the angle (in degrees) between the shuttle's velocity and the Earth's field?

Question

contestada

Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2020-05-16T02:16:58+02:00

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is [tex]\theta = 24.2^o[/tex]

Explanation:

From the question we are told that

The length of eire let out is [tex]L = 250 \ m[/tex]

The emf generated is [tex]\epsilon = 40 V[/tex]

The earth magnetic field is [tex]B = 5.0 *10^{-5} T[/tex]

The speed of the shuttle and tether is [tex]v = 7.80 * 10^3 \ m/s[/tex]

The emf generated is mathematically represented as

[tex]\epsilon = L\ v\ B\ sin \ \theta[/tex]

making [tex]\theta[/tex] the subject of the formula

[tex]\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ][/tex]

substituting values

[tex]\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ][/tex]

[tex]\theta = 24.2^o[/tex]