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34kurt
3 years ago
14

During electrical storms, a bolt of lightning can transfer 10 C of charge in 2.0 µs (the amount of time can vary considerably).

We can model such a bolt as a very long current carrying wire. a. What is the magnetic field 1.0 m from such a bolt? What is the field at 1.0 km away? b. How do the field compare with Earth’s magnetic field?
Physics
1 answer:
nydimaria [60]3 years ago
4 0

Answer:

Explanation:

charge, q = 10 C

time, t = 2 micro second

Current, i = q / t

i = 10 / (2 x 10^-6) = 5 x 10^6 A

(a)

distance, d = 1 m

the formula for the magnetic field is given by

B = \frac{\mu_{0}}{4\pi}\frac{2i}{d}

B = 10^{-7}\frac{2\times 5\times 10^{6}}{1}

B = 1 Tesla

Now the distance is d' = 1 km = 1000 m

B' = \frac{\mu_{0}}{4\pi}\frac{2i}{d}

B' = 10^{-7}\frac{2\times 5\times 10^{6}}{1000}

B' = 0.001 Tesla

(b) The magnetic field of earth is Bo = 3 x 10^-5 tesla

B / Bo = 3.3 x 10^4

B'/Bo = 33.3

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3 years ago
A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of
kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

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The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

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(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0
3 years ago
Calculate the force experienced by a positive test charge of 8.1 × 10-9 coulombs if the electric field strength is 9.4 × 107 new
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The electric force exerted by an electric field of intensity E on a charge q is equal to the product between E and q, so:
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3 years ago
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