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Aleksandr-060686 [28]
2 years ago
15

It is determined that a certain light wave has a wavelength of 3.012 x 10-12 m. The light travels at 2.99 x 108 m/s. What is the

frequency of the light wave? (Round your answer to three significant figures.)
*Answer in HZ units
Physics
1 answer:
gayaneshka [121]2 years ago
3 0

Answer:

id.k

Explanation:

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If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni
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Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

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Which simple machine belongs to the inclined plane family
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The wedge and screw simple machines
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Which of the following statements is true about a current-carrying wire in a magnetic field? A. Reversing the current direction
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A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horiz
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The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

B(h) = B(coil) = μ₀ NI / 2R

B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3

B(h) = 12.6 μT

Learn more about magnetic field here:

brainly.com/question/14411049

#SPJ4

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1 year ago
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