Ask question

Ask question

All categories

3 years ago

8

Which of the following is a characteristic of electromagnetic waves?

2 answers:

OleMash [197]3 years ago

6 0

The answer is B. They have a dual wave-particle nature.

fomenos3 years ago

4 0

Answer:

They have a dual wave-particle nature.

Explanation:

Electromagnetic waves consist of periodic oscillations of electric and magnetic field in a plane perpendicular to the direction of motion of the wave (in fact, they are also classified as transverse waves).

Electromagnetic waves have a wave nature, however they also have particle nature - in fact, it has been proved in some experiment (e.g. photoelectric effect) that in some conditions they act as packets of particles - called photons. Therefore, the option

They have a dual wave-particle nature.

is correct.

Other options are wrong because:

They are all invisible. --> False because visible light (which is part of the electromagnetic spectrum, so they are electromagnetic waves) is visible

They can only travel without a medium. --> False because they can also travel in a vacuum

They are slower than sound waves. --> False because they travel much faster (they travel at the speed of light in a vacuum, $c=3.0\cdot 10^8 m/s$ , while sound travels at 343 m/s in air, for instance)

You might be interested in

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

vazorg [7]

Answer: 1.907

Explanation:

I did the math

3 0

4 years ago

The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.

MAVERICK [17]

The correc answer is B.

7 0

3 years ago

How many seconds will it take for a satellite to travel 450,000 m at a rate of 120 m/s?

V125BC [204]

3750 seconds to travel that far

5 0

3 years ago

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

3 years ago

Read 2 more answers