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sukhopar [10]
4 years ago
11

In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 3.42 x 104-m length of wire

was let out by the space shuttle Atlantis to generate a motional emf. The shuttle had an orbital speed of 6.80 x 103 m/s, and the magnitude of the earth's magnetic field at the location of the wire was 6.56 x 10-5 T. If the wire had moved perpendicular to the earth's magnetic field, what would have been the motional emf generated between the ends of the wire
Physics
1 answer:
makvit [3.9K]4 years ago
8 0

Answer:

15116.4 V

Explanation:

B = Earth's magnetic field at the location = 6.5\times 10^{-5}\ T

L = Length of the wire = 3.42\times 10^4\ m

v = Velocity of the shuttle = 6.8\times 10^8\ m/s

The emf generated is given by

E=BLv\\\Rightarrow E=6.5\times 10^{-5}\times 3.42\times 10^{4}\times 6.8\times 10^3\\\Rightarrow E=15116.4\ V

The motional emf generated between the ends of the wire is 15116.4 V

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Answer:

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Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

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Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

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v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

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An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

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3 years ago
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A sphere of radius 5.00 cmcm carries charge 3.00 nCnC. Calculate the electric-field magnitude at a distance 4.00 cmcm from the c
kogti [31]

Answer:

a) E = 8628.23 N/C

b) E = 7489.785 N/C

Explanation:

a) Given

R = 5.00 cm = 0.05 m

Q = 3.00 nC = 3*10⁻⁹ C

ε₀ = 8.854*10⁻¹² C²/(N*m²)

r = 4.00 cm = 0.04 m

We can apply the equation

E = Qenc/(ε₀*A)  (i)

where

Qenc = (Vr/V)*Q

If    Vr = (4/3)*π*r³  and  V = (4/3)*π*R³

Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³

then

Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C

We get A as follows

A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²

Using the equation (i)

E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)

E = 8628.23 N/C

b) We apply the equation

E = Q/(ε₀*A)  (ii)

where

r = 0.06 m

A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²

Using the equation (ii)

E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)

E = 7489.785 N/C

6 0
3 years ago
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