Physics College

Which of these is a property of an electromagnetic wave? A)magnetic and electric fields oscillate perpendicular to each other but not to the velocity of the wave B)transports energy C)has a magnetic wave but no electric wave

Answers

Answer 1

Electromagnetic waves :

- are transverse waves

- Can travel through a vacuum

- Tranport energy from one place to another

- can be reflected

-can be refracted

Correct options:

B)transports energy

Related Questions

What is the initial velocity of an automobile acquiring a final velocity of 32 m/s with an acceleration of 4.0m/s ²

Answers

Answer:

Explanation:

Given:

V = 32 m/s

a = 4.0 m/s²

__________

V₀ - ?

V = V₀ + a*t

V₀ = V - a*t = 32 - 4*t

Time is not set according to the condition of the problem!

There's not enough given information t o answer the question. It depends on how long the car has been accelerating.

it could be 28 m/s 1 second ago.

it could be 16 m/s 4 seconds ago.

it could be 10 m/s 5.5 seconds ago.

etc.

i'll take a wild guess and speculate that the question actually tells how long the car has been accelerating, but you didn't copy that part.

An archery bow is drawn a distance d = 0.39 m and loaded with an arrow of mass m = 0.088 kg. The bow acts as a spring with a spring constant of k = 195 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow. If the arrow is shot at an angle of θ = 45° above the horizontal, how high, in meters above the initial height, will the arrow be when it reaches its peak?

Answers

The maximum height reached by the arrows is determined as 8.6 m.

What is the initial speed of the arrow?

The initial velocity of the arrow is calculated by applying the principle of conservation of energy as shown below;

K.E = U

where;

K.E is the kinetic energy of the arrowU is the elastic potential energy of the bow

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v = √(kx²/m)

where;

k is spring constant of the bowm is the mass of the arrowx is the extension of the bow

v = √(195 x 0.39²/0.088)

v = 18.36 m/s

The maximum height reached by the arrow is calculated as follows;

H = (v² sin²θ) / (2g)

where;

θ is angle of projection of the arrowg is acceleration due to gravity

H = (18.36² (sin45)²) / (2 x 9.8)

H = 8.6 m

Thus, the height of the arrow above the ground when it reaches its peak is 8.6 m.

Learn more about maximum height here: https://brainly.com/question/12446886

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A steel cable on a bridge has a linear mass density of 15 kg/m. If the cable has been pulled taunt with a tension of 5536 N, what is the speed of a wave on it?

Answers

[tex]\begin{gathered} T=5,536N \\ \mu=15\text{ kg/m} \\ v=? \\ v=\sqrt{\frac{T}{\mu}} \\ v=\sqrt{\frac{5,536N}{15\text{ kg/m}}} \\ v=19.21\text{ m/s} \\ The\text{ speed of a wave on the cable is 19.21m/s} \end{gathered}[/tex]

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