Carbon dioxide is a gas produced by___ and used by __

How are wave properties and energy related?<br> give me examples

klasskru [66]

Answer:

They don’t ‘represent’ anything, they are properties of the wave.

Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.

If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.

For many waves, we don’t perceive them at all (e.g. radio waves).

For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.

6 0

2 years ago

A runner exerts a net force of 335 n to accelerate at a rate of 2.5 m/s what is the runners mass

Olegator [25]

F=ma
M=F/a
M=335/2.5
M=134 kg

8 0

2 years ago

What is the study of energy and how it affects matter called?

bogdanovich [222]

Physics

I hope I helped:)

6 0

2 years ago

Which of the following terms refer to the area immediately around the crime?

UkoKoshka [18]

Answer:

A....It is a crime sene

Explanation:

6 0

2 years ago

Read 2 more answers

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$