A physics student notices that 4.0 waves arrive at a beach every 20 seconds. The frequency of the wave is

Nookie1986 [14]

Answer:

Objects; waves.

Explanation:

Waves interact with objects and other waves. Thus, waves are used on objects such as mobile phones and can be transformed from one form to another.

There are various types of waves in our physical environment such as gamma rays, x-rays, ultraviolet waves, radio waves etc.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.

Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data. Some examples of communication technologies that uses radio waves are radio set, mobile phones, television etc.

6 0

3 years ago

If velocity is 0.2m/s, what is the kinetic energy?

Natalija [7]

$\huge\boxed{♧ \: \mathfrak{ \underline{Answer} \: ♧}}$

we know,

$\boxed{kinetic \: \: energy = \frac{1}{2} m {v}^{2} }$

So,

$\longmapsto \dfrac{1}{2} \times 10 \times 0.2 \times 0.2$

$\longmapsto0.2 \: joules$

3 0

4 years ago

What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?

zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0

3 years ago

Read 2 more answers

Solve for (b) how many revolutions it takes for the cd to reach its maximum angular velocity in 1.36s?

Digiron [165]

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

$\omega = 30 rad/s$

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

$\theta = (\frac{\omega_0 + \omega}{2})t$

where:

$\theta$ is the angular displacement of the cd during the time interval t

$\omega_0$ is the initial angular velocity of the cd

$\omega$ is the final angular velocity

Here we have:

t = 1.36 s

$\omega_0 = 0$ (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

$\theta=(\frac{0+30}{2})(1.36)=20.4 rad$

And since $1 rev = 2 \pi rad$ , we can convert into number of revolutions completed:

$\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev$

Learn more about angular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

8 0

3 years ago

A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin

jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: $F_y=mg+N$ [/tex]

c) The velocity at which the roller coaster will fall is $[tex]v=\sqrt{gr}$ [/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

The weight of the roller coaster, acting downward, indicated by $mg$ (where m is the mass of the roller coaster and g is the acceleration of gravity)
The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

$F_y = mg + N$

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

$m\frac{v^2}{r}=mg+N$

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

$m\frac{v^2}{r}=mg+N$

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

$v=\sqrt{gr+\frac{Nr}{m}}$

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

$v=\sqrt{gr}$

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

$g=9.8 m/s^2$

We said that the minimum velocity that the cart must have in order not to fall at the top is

$v=\sqrt{gr}$

And substituting, we find

$v=\sqrt{(9.8)(30)}=17.1 m/s$

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

$mgh = \frac{1}{2}mv^2 + mg(2r)$