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Ahat [919]
3 years ago
10

Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, stra

ight wire. If you were unfortunate enough to be 4.9 m away from such a lightning bolt, how large a magnetic field would you experience
Physics
2 answers:
Tamiku [17]3 years ago
6 0

Answer: 8.16*10^-4 T

Explanation:

Given

Current of the lightening bolt, I = 20 kA

Distance from the strike of the lightening bolt, r = 4.9 m

To solve, we use the formula

B = [μ(0) * I] / 2πr, where

B = magnetic field of the lightening

μ = permeability constant = 4π*10^-7 N/A²

I = current of the lightening

r = distance from the lightening strike

B = [(4 * 3.142*10^-7) * 20*10^3] / (2 * 3.142 * 4.9)

B = (12.568*10^-7 * 20*10^3) / 6.284 * 4.9

B = 0.025 / 30.79

B = 8.16*10^-4 T

The magnetic field to be experienced would be 8.16*10^-4 T large

GalinKa [24]3 years ago
4 0

Answer:

how large a magnetic field would you experience = 8.16 x 10∧-4T

Explanation:

I = 20KA = 20,000A

r = 4.9 m

how large a magnetic field would you experience =  u.I/2πr

how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9

how large a magnetic field would you experience = 8.16 x 10∧-4T

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Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __
Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

ξ = Blv

Where,

ξ = Emf Induced

B = Magnetic Induction

l = Length of the conductor

v = Speed of the conductor.

As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

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3 years ago
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Heat rises therefore the heat from the fire rises up to your hand... i didnt have any answer choices to work with sorry
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3 years ago
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An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this
galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

\gamma = \frac{Mt}{J}

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

J = \frac{1}{2} * m * r^2

Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2

The wheel walls act like annular rings, for these the moment of inertia is:

J = \frac{1}{2} * m * (re^2 - ri^2)

Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

J = m * r^2

Jtread = 10 * 0.33^2 = 1.09 kg*m^2

The axle acts like a rod, which is the same as the disk:

Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2

The drive shaft acts like a rod too:

Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}

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