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3 years ago

Which are properties of an electromagnetic wave?

2 answers:

8 0

*(In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - the fastest speed possible.*)

anastassius [24]3 years ago

7 0

-Electromagnetic waves carry transverse vibrations in electrical and magnetic fields, not vibrating particles.
-Electromagnetic waves do not need matter to travel through - they can travel through empty space (a vacuum).
-In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - the fastest speed possible.
-When electromagnetic waves travel through matter (for example, light through air or glass), they travel a bit slower than this but rarely less than half as fast as in a vacuum. The value for light travelling through an optical fibre, for example, is taken as 2 x 108 m/s.

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An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion

anyanavicka [17]

Answer:

attached below is the free body diagram of the missing illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

$E_{e} = E_{3} - E_{1}$

E $_{e}$ = initial kinetic energy of the electron

E $_{1}$ = -4 eV

E $_{3}$ = -1 eV

insert the values into the equation above

$E_{e}$ = -1 -(-4) eV

= -1 + 4 = 3 eV

6 0

3 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

so above is the instantaneous accelerations

7 0

3 years ago

Energy from the sun drives the _ of the water from the ocean surface. As the air moves up the west side of the mountain, the wat

marin [14]

Answer: The air is full of water, as water vapor, even if you can't see it. Condensation is the process of water vapor turning back into liquid water, with the best example being those big, fluffy clouds floating over your head. And when the water droplets in clouds combine, they become heavy enough to form raindrops to rain down onto your head.

Explanation:

6 0

2 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

3 years ago

More buffer zones would be needed around a large protected area vs. several patches of protected area of equal square footage.

Fudgin [204]

Your answer is it is True. Please no mean comments I'm new to brainly

6 0

3 years ago

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