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Nezavi [6.7K]
3 years ago
12

Given that the frequency of an EM wave is 4THz, what is its wavelength?

Physics
1 answer:
Vilka [71]3 years ago
8 0

Answer:

The wavelength of the EM wave is 7.5 * 10⁻⁴ m

Explanation:

The velocity of a wave is related to its wavelength by the following formula;

velocity = wavelength * frequency

For an electromagnetic (EM) wave, its velocity is equal to the velocity of light, c = 3.0 * 10⁸ m/s

Given that the frequency and veloity of the given EM wave in the question is known, its wavelength is calculated as follows:

wavelength = velocity/frequency

where velocity of the EM wave = 3.0 * 10⁸ m/s;

frequency = 4THz = 4 * 10¹² Hz

wavelength = 3.0 * 10⁸m/s / 4 * 10¹² Hz

wavelength = 7.5 * 10⁻⁴ m

Therefore, the wavelength of the EM wave is 7.5 * 10⁻⁴ m

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A nonconducting ring of radius R is uniformly charged with a total positive charge q. The ring rotates at a constant angular spe
kipiarov [429]

The magnitude of the magnetic field on the axis of the ring (1/2)R from its center is [ μ₀ ωqR² ] / [2.5(√5)πR ].

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 radius of the nonconducting ring is R.

The ring is uniformly charged q.

The angular speed of the ring is ω.

The ring is x = (1/2)R from the center of the ring.

The magnetic field on the axis of a current loop is given as:

B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]

Now, I = q / [2π/ω]

When x = R/2 the magnitude of the magnetic field is:

B = [ μ₀ ωqR² ] / [4π( x² + R²)^{3/2} ]

B = [ μ₀ ωqR² ] / [4π( ( R/2 )² + R² )^{3/2} ]

B = [ μ₀ ωqR² ] / [4π( [5/4]R² )^{3/2} ]

B = [ μ₀ ωqR² ] / [2.5(√5)πR ]

Learn more about the magnetic field here:

brainly.com/question/14411049

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6 0
2 years ago
A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?
Amanda [17]

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

4 0
3 years ago
Texting and driving is a dangerous act
sveta [45]

Answer:

s = 92.36 m

Explanation:

The distance covered by an object moving with uniform speed is given by the following formula:

s = vt

where,

s = distance covered by car during the time when driver was distracted in texting = ?

v = uniform speed of car = (95 km/h)(1 h/3600 s)(1000 m/1 km) = 26.39 m/s

t = time taken by the driver for texting = 3.5 s

Therefore, using these values in equation, we get:

s = (26.39 m/s)(3.5 s)

<u>s = 92.36 m</u>

4 0
4 years ago
A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a po
morpeh [17]

Answer:

a) before immersion

C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F

q = CV = (1.68e-12)(255) = 4.28e-10 C

b) after immersion

q = 4.28e-10 C

Because the capacitor was disconnected before it was immersed, the charge remains the same.

c)*at 20° C

C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F

V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V

e)

U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J

U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J

ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J

8 0
3 years ago
To be Answered in Sentences...
Nikolay [14]

Answer:

1. The equivalent resistance for the combination of resistors in series is equal to the algebraic sum of all its individual resistances.

2. The Current will increase and causes it to have less restriction.

Explanation:

I did the Exam and made a 100% with these answers.

Hope this Helps!!!

8 0
3 years ago
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