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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:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

E = \dfrac{kq}{r^2}

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

q= \dfrac{Er^2}{k}

now,

q= \dfrac{180000\times 0.015^2}{9\times 10^9}

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2 years ago
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n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

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The measured value is 2.02 meters and the actual value is 2.00 meters.

V_{measured}=2.02\\\\V_{true}=2.00

Percent Error=\frac{2.02-2.00}{2.00} *100

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Percent Error=\frac{0.02}{2.00}*100

Next, divide 0.02 by 2.00

PercentError=0.01 *100

Finally, multiply 0.01 and 100.

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The percent error is 1%.

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2 years ago
An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable i
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1 year ago
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Hi!


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2 years ago
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