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Makovka662 [10]
2 years ago
9

It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica

tions, 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?

Physics
1 answer:
denis23 [38]2 years ago
6 0

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  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

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

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

The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

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

substituting values

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

                        \theta =   24.2^o

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Answer:

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2 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
2 years ago
A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does
astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus =  14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

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Bulk modulus is equal to

B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }

now

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }

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change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

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1 year ago
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