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Deffense [45]
3 years ago
7

We can model a lightning bolt as a very long, straight wire. If a lightning bolt carries a current of 30 kA, and you are unfortu

nate enough to be 8.9 m away from the lightning bolt, how large of a magnetic field do you experience
Physics
1 answer:
Liula [17]3 years ago
8 0

Answer:

Magnetic field experienced = 4.5 × 10⁻⁴ T

Explanation:

The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by

B = (μ₀I)/(2πr)

B = ?

I = 20 KA = 20000 A

r = 8.9 m

μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A

B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T

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Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

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3 years ago
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How does freezing and evaporation affect salinity?
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The answer is C, because the water molecules are evaporating and the salt molecules are staying the same.
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2 years ago
A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies
Roman55 [17]

Answer:

The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

3 0
3 years ago
Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon
Mice21 [21]

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

6 0
2 years ago
The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
Gemiola [76]

Answer:

6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

I=0.040 \mu A=4\cdot 10^{-8}A

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T

5 0
3 years ago
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