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arsen [322]
3 years ago
8

Which of the following statements are true regarding electromagnetic waves traveling through a vacuum? (Select all that apply.)

All waves have the same wavelength. All waves have the same frequency. All waves travel at 3.00 108 m/s. The electric and magnetic fields associated with the waves are perpendicular to each other and to the direction of wave propagation. The speed of the waves depends on their frequency.
Physics
2 answers:
Ber [7]3 years ago
4 0

Correct choices:

- All waves travel at 3.00 108 m/s.

- The electric and magnetic fields associated with the waves are perpendicular to each other and to the direction of wave propagation.

Explanation:

Let's analyze each statement:

- All waves have the same wavelength. --> FALSE. Electromagnetic waves have a wide range of wavelengths, from less than 10 picometers (gamma rays) to hundreds of kilometers (radio waves)

- All waves have the same frequency. --> FALSE. As for the wavelength, electromagnetic waves have a wide range of frequencies also.

- All waves travel at 3.00 108 m/s. --> TRUE. This value is called speed of light, and it is one of the fundamental constant: it is the value of the speed of all electromagnetic waves in a vacuum.

- The electric and magnetic fields associated with the waves are perpendicular to each other and to the direction of wave propagation. --> TRUE. Electromagnetic waves are transverse waves, which means that their oscillations (represented by the electric field and the magnetic field) occurs perpendicularly to the direction of motion of the wave.

- The speed of the waves depends on their frequency. --> FALSE. In a vacuum, the speed of ALL electromagnetic waves is always equal to c, regardless of the frequency.

kotegsom [21]3 years ago
3 0

Answer:

option C and D

Explanation:

Electromagnetic waves can travel in vacuum as well as in a medium. The different waves have different frequency and wavelength but have same speed in vacuum (3.00 x 10⁸ m/s).

These waves carry the energy via oscillating electric and magnetic fields. The electric and magnetic fields oscillate perpendicular to each other and to the direction of motion of the wave.

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jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

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The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

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3 years ago
Please please help me please please help please please help me please
Kaylis [27]

Answer:

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So, Basically, it is B I believe.

Hope It Helps!

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